SAT Math: Everything You Need to Know
As a standardized test, College Board's SAT is designed to be consistent, fair, and predictable for all test-takers. While this makes preparing for and taking the test painfully boring, it also ensures that the questions always focus on the same core concepts, regardless of the specific test. These concepts are limited to material typically covered in 9th through 11th grade (think algebra, geometry, trigonometry, and basic statistics and data analysis), meaning no calculus, linear algebra, or other more advanced topics. As a result, the SAT is more about speed and accuracy than sheer difficulty. Luckily, Desmos updates as fast as you can type and never makes mistakes, making it your best friend for saving time and minimizing those pesky computational errors.
This lesson assumes that you have a working knowledge of the fundamentals presented in Desmos101. This is not a traditional math course, so you should already be familiar with the underlying math concepts.
Key Concepts
Shortcuts •
Useful Functions •
Single Variable Equations •
Systems of Equations •
Systems of Inequalities •
Equivalent Expressions •
Number of Solutions •
Graph Features •
Composite Functions •
Geometry •
Sliders •
Tables & Regressions •
General SAT Advice
Keyboard Shortcuts
Here are a few keyboard shortcuts for Desmos that may save you some precious time on test day:
Function
\(a^b\)
\(a_b\)
\(\leq\) or \(\geq\)
\(a\cdot b\)
\(\frac{a}{b}\)
\(\sqrt{a}\)
\(\left|a\right|\)
\(\sin\)
\(\cos\)
\(\tan\)
\(\pi\)
Shortcut
^ (Shift + 6)
_ (Shift + -)
<= or >=
*
/
sqrt
|
sin
cos
tan
pi
Want to see the full list of keyboard shortcuts? Click here.
You certainly don't need to memorize all these shortcuts, but they can be helpful to speed up your workflow in Desmos. As you use Desmos more, you'll naturally pick up the ones that are most useful to you. Try retyping the expressions on lines 2, 4, and 6 of the expression list without the Desmos keypad.
Practice Question #1
\[y=\sqrt{x^{\frac{4}{\pi}}}\]
How would you type this expression into Desmos using your keyboard?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. Both \(y\) and \(=\) can be represented by standard keyboard characters. The square root can be entered using the sqrt shortcut, which is much faster than using the Desmos keypad. You can exponentiate \(x\) using the caret (^), which is also a standard keyboard character. Finally, entering the numerator, can can create the fraction with a standard forward slash and use the pi shortcut to enter the \(\pi\) in the denominator.
Answer A is incorrect because it uses symbols that cannot be entered directly in Desmos without using the Desmos keypad (which is slow!).
Answer B is incorrect because it uses the wrong keyboard shortcut for the square root. The correct shortcut is sqrt, not squareroot.
Answer D actually produces the correct graph, but uses extra parentheses around the exponent which are unnecessary and not the most efficient way to enter the equation in Desmos. The SAT rewards efficiency, so the best answer is C, which uses the fewest keystrokes while accurately representing the given equation.
Not quite...
Show Answer
The correct answer is C. Both \(y\) and \(=\) can be represented by standard keyboard characters. The square root can be entered using the sqrt shortcut, which is much faster than using the Desmos keypad. You can exponentiate \(x\) using the caret (^), which is also a standard keyboard character. Finally, entering the numerator, can can create the fraction with a standard forward slash and use the pi shortcut to enter the \(\pi\) in the denominator.
Answer A is incorrect because it uses symbols that cannot be entered directly in Desmos without using the Desmos keypad (which is slow!).
Answer B is incorrect because it uses an asterisk instead of a caret. This produces \(y=\sqrt{x\cdot{\frac{4}{\pi}}}\) instead of \(y=\sqrt{x^{\frac{4}{\pi}}}\).
Answer D actually produces the correct graph, but uses extra parentheses around the exponent which are unnecessary and not the most efficient way to enter the equation in Desmos. The SAT rewards efficiency, so the best answer is C, which uses the fewest keystrokes while accurately representing the given equation.
Useful Functions & Timesaves
Mean & Median
You can find the mean and median of a set of numbers using the mean and median functions, as shown on lines 1-2 of the expression list. Just type "mean" or "median," enter the values of your set, and Desmos does the rest. If there are multiple instances of the same value in the set, make sure to include them.
Select All (For Expressions)
If you want to enclose an entire expression in parentheses, do Ctrl/Cmd + A to select everything, then ( or ) to add the parentheses. This is useful if you have a complicated expression that you want to quickly apply an operation to without having to retype the entire expression. Try enclosing the expression on line 4 of the expression list in parentheses and raising it to the power of \(\pi\).
Decimals ↔ Fractions
If you have an expression that returns a fractional value, you can toggle between decimals and fractions by clicking the fraction icon to the left of the corresponding line in the expression list. Try finding the simplest fractional form of the expression on line 6 of the expression list using the fraction/decimal toggle.
Omittable Variables
While it's good practice to include the dependent variable in your equations, you can omit it for single-variable equations if you just want to see the graph. For example, instead of typing \(y = 2x + 3\), you can simply type \(2x + 3\) and Desmos will automatically assume that the dependent variable is \(y\). Try clicking between the expressions on lines 8-9 of the expression list to verify that they represent the same graph.
Quick Nav
To quickly zoom in and out your graph view, scroll while hovering over the graph. To change the domain/range of a particular axis, hold Shift while dragging the axis you want to resize.
Want to see the full list of supported functions? Click here.
Practice Question #2
Given the set \(S = \{5,3,2,1,7,0\}\), what is the ratio of the median to the mean of the set, expressed in simplest fractional form? Try to solve this problem with Desmos as you would on the SAT.
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is A. We can use the built-in Desmos functions to find that the median and mean of set \(S\) are 2.5 and 3, respectively. The ratio of the median to the mean is thus \(\frac{2.5}{3}\). To express this in simplest form, we can simply let Desmos do the work for us by clicking the fraction icon to the left of the expression. This gives us \(\frac{5}{6}\).
Note that this whole process can be done in a single line in the expression list: \(\frac{\text{median}(5,3,2,1,7,0)}{\text{mean}(5,3,2,1,7,0)}\). This is a great example of how Desmos can save you time and effort on the SAT. In fact, this could be done even faster using lists, which we will cover in Secret Tips + Artwork.
Answer B is incorrect because it is the reciprocal of the correct answer. This is the ratio of the mean to the median, not the median to the mean.
Answer C is incorrect because it is a rounded decimal representation of the correct answer, and the question specifically asks for the answer in simplest fractional form. Note that the SAT allows decimal answers, but they must be truncated or rounded at the fourth digit. This answer is rounded at the third digit, which would actually be marked incorrect on the SAT!
Answer D is incorrect because it is the decimal representation of the reciprocal of the correct answer. This is the ratio of the mean to the median, not the median to the mean, and is not in simplest fractional form.
Not quite...
Show Answer
The correct answer is A. We can use the built-in Desmos functions to find that the median and mean of set \(S\) are 2.5 and 3, respectively. The ratio of the median to the mean is thus \(\frac{2.5}{3}\). To express this in simplest form, we can simply let Desmos do the work for us by clicking the fraction icon to the left of the expression. This gives us \(\frac{5}{6}\).
Note that this whole process can be done in a single line in the expression list: \(\frac{\text{median}(5,3,2,1,7,0)}{\text{mean}(5,3,2,1,7,0)}\). This is a great example of how Desmos can save you time and effort on the SAT. In fact, this could be done even faster using lists, which we will cover in Secret Tips + Artwork.
Answer B is incorrect because it is the reciprocal of the correct answer. This is the ratio of the mean to the median, not the median to the mean.
Answer C is incorrect because it is a rounded decimal representation of the correct answer, and the question specifically asks for the answer in simplest fractional form. Note that the SAT allows decimal answers, but they must be truncated or rounded at the fourth digit. This answer is rounded at the third digit, which would actually be marked incorrect on the SAT!
Answer D is incorrect because it is the decimal representation of the reciprocal of the correct answer. This is the ratio of the mean to the median, not the median to the mean, and is not in simplest fractional form.
Single Variable Equations
Solving Single Variable Equations by Graphing
Virtually any single variable equation (most often where you need to isolate or solve for \(x\)) can be solved in Desmos by simply typing the equation into the expression list. For example, try typing the equation \(\frac{55}{x+6}=x\) into the expression list. If you zoom out (which is always a good idea when graphing equations, especially since the SAT likes to give equations with solutions of varying magnitudes), you should see two vertical lines at \(x = -11\) and \(x = 5\). These are the solutions to the equation.
Arranging Single Variable Equations as Systems of Equations
You may notice that Desmos does not allow you to select the vertical lines, which can be problematic for non-integer solutions. To find the exact solution, you can graph each side of the equation as a function of \(x\), and then find the \(x\)-coordinate(s) of where the two functions intersect. This is a great way to understand conceptually what is happening when we solve for a single variable equation and is a good way to verify your solution. Try graphing the two sides of the equation \(\frac{55}{x+6}=x\) as functions of \(x\) and identifying the points of intersection.
Additional Steps
Occasionally the SAT will ask you to do something else with the solution(s) to a single variable equation. Here is a short list of common things you may be asked to do with the solution(s):
- Interpret the solution in context
- Choose the positive/negative or smallest/largest solution
- Plug the solution into another expression
- Find the sum or product of solutions
- Find the number of solutions (more on scenarios with zero/infinite solutions later!)
These are all pretty straightforward once you have the solutions themselves, but pay attention to exactly what the question is asking.
Try finding the sum of the solutions to the equation \(\frac{55}{x+6}=x\).
Practice Question #3
If \(2(3t - 10) + t = 40 + 4t\), what is the value of \(3t\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is B. We can graph the equation by typing it into the expression list, substituting \(x\) for all instances of \(t\) because Desmos only supports implicit functions of \(x\) and \(y\). Zooming out, we find a vertical line at \(x = 20\). Thus, the solution to the equation is \(t = 20\), so the value of \(3t\) is \(3(20) = 60\).
Answer A is incorrect. This result stem from forgetting the "\(+ t\)" term on the left-hand side of the equation, in which case the equation yields a solution of \(t = 30\).
Answer C is incorrect. This is the value of \(t\), not \(3t\). Make sure to be careful about what the question is asking!
Answer D is incorrect and may result from conceptual or calculation errors. This is not the value of \(3t\).
Not quite...
Show Answer
The correct answer is B. We can solve the equation by typing it into the expression list, substituting \(x\) for all instances of \(t\) because Desmos only supports implicit functions of \(x\) and \(y\). Zooming out, we find a vertical line at \(x = 20\). Thus, the solution to the equation is \(t = 20\), so the value of \(3t\) is \(3(20) = 60\).
Answer A is incorrect. This result stem from forgetting the "\(+ t\)" term on the left-hand side of the equation, in which case the equation yields a solution of \(t = 30\).
Answer C is incorrect. This is the value of \(t\), not \(3t\). Make sure to be careful about what the question is asking!
Answer D is incorrect and may result from conceptual or calculation errors.
Systems of Equations
Graphing Systems of Equations
Similar to single variable equations, virtually any system of equations can be solved in Desmos simply by typing the equations into the expression list and seeing where they intersect. Try typing the system of equations \(y = 2x + 3\) and \(11 = 2y + x\) into the expression list. You should see a point of intersection at \((1,5)\), which is the solution to the system of equations. If you want to save this point for later, you can hover over the point of intersection and click the export icon to the right of the coordinate to save it to the expression list.
Conceptual Interpretation
Mathematically, you can think of the first line as every point that satisfies the equation \(y = 2x + 3\), and the second line as every point that satisfies the equation \(11 = 2y + x\). The point of intersection is the point that satisfies both equations. This is true for systems of equations of any form (quadratic, exponential, etc.) as long as you have two equations and two variables. Sometimes there may be more than one point of intersection, or no points of intersection at all. This goes back what we covered on page 4 about single variable equations. Any single variable equation can be arranged as a system of equations, and vice versa.
Practice Question #4
\[8x + y = -11\]
\[2x^2 = y + 341\]
The graphs of the equations in the given system of equations intersect at the point \((x, y)\) in the \(xy\)-plane. What is a possible value of \(x\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is D. We can graph the two equations by typing them into the expression list. Zooming out quite a bit (something you should get used to for the SAT), we find that the two equations intersect at \((-15,109)\) and \((11,-99)\). We're asked to find a possible value of \(x\) for an intersection \((x,y)\). Both \(x\)-coordinates meet this criterion, but only \(-15\) is one of the answer choices, so the correct answer is D.
Answer A is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Answer B is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Answer C is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Not quite...
Show Answer
The correct answer is D. We can graph the two equations by typing them into the expression list. Zooming out quite a bit (something you should get used to for the SAT), we find that the two equations intersect at \((-15,109)\) and \((11,-99)\). We're asked to find a possible value of \(x\) for an intersection \((x,y)\). Both \(x\)-coordinates meet this criterion, but only \(-15\) is one of the answer choices, so the correct answer is D.
Answer A is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Answer B is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Answer C is incorrect and may result from conceptual or calculation errors. This is not a solution to the given system of equations.
Systems of Inequalities
Graphing Systems of Inequalities
You can also use a similar method for solving systems of inequalities to find the solution to a system of inequalities. Try typing the system of inequalities \(y \leq 2x + 3\) and \(11 \geq 2y + x\) into the expression list. You should see two shaded regions that satisfy each inequality. Remember from Desmos 101 that any point where both regions overlap is a solution to the system of inequalities.
Coordinate Tables
Sometimes the SAT will give you a series of coordinate tables and ask you to identify the table that represents the system of inequalities. Simply graph the inequalities and see which table contains only points within the overlapping colored region. If you are given a table of points, you can also create a table by typing "table" in an empty expression line and entering the points into the table. This is a great way to visualize the points and see which ones are solutions to the system of inequalities. We will talk more about tables in later sections. Try creating a table with the points \((2,-3)\), \((0,3)\), \((1,5)\), and \((-4, -7.5)\) and verify that they are all solutions to the system of inequalities.
Practice Question #5
\[y > 4x - 6\]
\[4y \leq -\frac{1}{2}x + 1\]
For which of the following sets are all the values of \(x\) and their corresponding values of \(y\) solutions to the given system of inequalities?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. We can graph the two inequalities by typing them into the expression list. If we make a table for each answer choice and insert the corresponding points, we find that the only set of points for which all the points fall within the overlapping colored region is C. Thus, all of the values of \(x\) and their corresponding values of \(y\) in answer choice C are solutions to the given system of inequalities.
Answer A is incorrect because the points \((-6, 2)\) and \((2, 0)\) do not fall within the overlapping colored region and are therefore not solutions to the given system of inequalities.
Answer B is incorrect because the point \((0, -6)) does not fall within the overlapping colored region and is therefore not a solution to the given system of inequalities. Remember that dotted lines indicate that the inequality is non-inclusive, so the point does not satisfy the first inequality.
Answer D is incorrect because the points \((-3, 1)\) and \((6, 7)\) do not fall within the overlapping colored region and are therefore not solutions to the given system of inequalities.
Not quite...
Show Answer
The correct answer is C. We can graph the two inequalities by typing them into the expression list. If we make a table for each answer choice and insert the corresponding points, we find that the only set of points for which all the points fall within the overlapping colored region is C. Thus, all of the values of \(x\) and their corresponding values of \(y\) in answer choice C are solutions to the given system of inequalities.
Answer A is incorrect because the points \((-6, 2)\) and \((2, 0)\) do not fall within the overlapping colored region and are therefore not solutions to the given system of inequalities.
Answer B is incorrect because the point \((0, -6)) does not fall within the overlapping colored region and is therefore not a solution to the given system of inequalities. Remember that dotted lines indicate that the inequality is non-inclusive, so the point does not satisfy the first inequality.
Answer D is incorrect because the points \((-3, 1)\) and \((6, 7)\) do not fall within the overlapping colored region and are therefore not solutions to the given system of inequalities.
Equivalent Expressions Part 1: Explicit Expressions
The SAT will often provide an expression and then ask you to identify the equivalent expression out of the four answer choices. To do this, first graph the given expression and then graph each answer choice. The equivalent expression is the one that produces the same graph.
Try graphing the expression \(\frac{4}{4x - 5} - \frac{1}{x + 1}\) and each of the potentially equivalent expressions:
- \(\frac{1}{(x + 1)(4x - 5)}\)
- \(\frac{3}{3x-6}\)
- \(\frac{9}{(x + 1)(4x - 5)}\)
- \(-\frac{1}{(x + 1)(4x - 5)}\)
You should find that \(\frac{9}{(x + 1)(4x - 5)}\) is the only expression that produces the same graph as the given expression. Click between these two expressions in the expression list to confirm that their graphs overlap.
You can use this method of plugging in potential expressions (or values, as we'll see next) to find an overlapping, equivalent for any type of function on the SAT. This will be especially useful when we talk about finding the number of solutions in future sections.
Practice Question #6
Which expression is equivalent to \(\frac{x^3 - 5x^2 + 6x}{x^2 - 4x}\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is B. We can graph the given expression and each of the answer choices by typing them into the expression list. We find that the only expression that produces the same graph as the given expression is the one in answer choice B. To confirm, we can click between the two expressions in the expression list to verify that their graphs overlap.
Answer A is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Answer C is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Answer D is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Not quite...
Show Answer
The correct answer is B. We can graph the given expression and each of the answer choices by typing them into the expression list. We find that the only expression that produces the same graph as the given expression is the one in answer choice B. To confirm, we can click between the two expressions in the expression list to verify that their graphs overlap.
Answer A is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Answer C is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Answer D is incorrect because its graph does not match the graph of the given expression. Thus, the two expressions are not equivalent.
Equivalent Expressions Part 2: Parameterized Expressions
Sometimes the SAT will give one explicit expression and one general expression with a parameter (often a letter like \(b\) or \(k\) which is defined as some integer), and ask you to find the value of the parameter that makes the two expressions equivalent. To do this, first graph the explicit expression and then graph the general expression.
Try graphing the explicit expression \(\frac{24}{6x + 42}\) and the general expression \(\frac{4}{x + b}\). You should see that Desmos gives you the option to create a slider for \(b\). You can create the slider automatically by making a new line or by accepting Desmos' suggestion.
Most of the time, the question will be multiple choice with four values for the parameter, so you can simply plug in each value and see which one produces the same graph as the explicit expression. Try plugging in the following values for \(b\) to see which one produces the same graph as the explicit expression:
- \(b = 7\)
- \(b = 10\)
- \(b = 24\)
- \(b = 252\)
If the question is free response, you can drag the slider to find the value of the parameter that produces the same graph as the explicit expression. We'll talk more about sliders later, but for these questions you don't need any additional functionality beyond the information provided in Desmos 101.
Practice Question #7
\[\sqrt{x^8y^9} = 5\]
The given expression is equivalent to \(\left(x^\frac{8}{9}y\right)^k = 5\), where \(k\) is a constant. What is the value of \(k\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is D. We can graph the given explicit expression and the parameterized expression by typing them into the expression list. We can make a slider for the parameter, \(k\), and change its value to each of the answer choices to see which one produces the same graph as the given expression. We find that the only value of \(k\) that produces the same graph as the given expression is D. To confirm, we can click between the two expressions in the expression list to verify that their graphs overlap.
Answer A is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Answer B is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Answer D is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Not quite...
Show Answer
The correct answer is D. We can graph the given explicit expression and the parameterized expression by typing them into the expression list. We can make a slider for the parameter, \(k\), and change its value to each of the answer choices to see which one produces the same graph as the given expression. We find that the only value of \(k\) that produces the same graph as the given expression is D. To confirm, we can click between the two expressions in the expression list to verify that their graphs overlap.
Answer A is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Answer B is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Answer D is incorrect because plugging in this value for \(k\) in the parameterized expression does not produce the same graph as the given explicit expression.
Number of (Real) Solutions Part 1: Single Variable Equations
Unless you see a quick algebraic solution (which is highly encouraged as a first attempt!), your first instinct when determining the number of solutions to a single variable equation should typically be to graph the equation in Desmos. As we have shown, this is a great way to visualize the solutions. Here are some ways to interpret the two primary results you may get from graphing a single variable equation:
A Positive Integer Number of Solutions
Recall that if you see one or more vertical lines, you can simply count the number of vertical lines to determine the number of solutions. Remember to zoom out to view the entire graph! Try graphing the equation \(x^2 + 3x - 4 = 0\) and count how many vertical lines (solutions) you see.
No Solution vs. Infinite Solutions
Unfortunately, if you don't see any vertical lines, you may have a little more work to do. On Desmos, both no solution and infinite solutions do not display anything for single variable equations. To determine which one you have, you should try to graph each side of the equation as a separate function of \(y\). If they overlap, each side is equal to the other for all values of \(x\) (as we know from the previous sections), so the equation has infinite solutions. If they do not overlap, most often when the sides of the equation represent two perfectly parallel lines, the equation has no solution. Sometimes the slopes of these lines are very similar and appear to overlap, but they are not exactly the same. In this case, try zooming out to see if the lines diverge.
Try graphing the equation \((x + 2)^2 = x^2 + 4x + 4\) and count how many vertical lines (solutions) you see. You should see that there are no vertical lines. To determine whether there are no solutions or infinite solutions, try graphing each side of the equation as a separate function of \(y\). You should see that their graphs overlap, so the equation has infinite solutions.
Practice Question #8
How many solutions does the equation \(10(15x - 9) = -15(6 - 10x)\) have?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is A. We can graph the equation by typing it into the expression list. We find that there are no vertical lines, so we need to determine whether there are no solutions or infinite solutions. To do this, we graph the two sides of the equation as separate functions of \(y\). We find that the two sides overlap, so the equation is true for all \(x\) values and has infinite solutions.
Answer B is incorrect because there are no vertical lines when we graph the equation.
Answer C is incorrect because there are no vertical lines when we graph the equation, and the two sides of the equation represent two lines (which cannot have two points of intersection).
Answer D is incorrect because the two sides of the equation overlap rather than being perfectly parallel. Therefore, the equation has infinite solutions.
Not quite...
Show Answer
The correct answer is A. We can graph the equation by typing it into the expression list. We find that there are no vertical lines, so we need to determine whether there are no solutions or infinite solutions. To do this, we graph the two sides of the equation as separate functions of \(y\). We find that the two sides overlap, so the equation is true for all \(x\) values and has infinite solutions.
Answer B is incorrect because there are no vertical lines when we graph the equation.
Answer C is incorrect because there are no vertical lines when we graph the equation, and the two sides of the equation represent two lines (which cannot have two points of intersection).
Answer D is incorrect because the two sides of the equation overlap rather than being perfectly parallel. Therefore, the equation has infinite solutions.
Number of (Real) Solutions Part 2: Systems of Linear Equations
For systems of two linear equations, it's always a good idea to start by graphing the two equations. As a general rule of thumb, you can tell how many solutions a system of equations has by looking at how many points of intersection it has.
If you see two lines that intersect at a single point, the system has one solution. If you see two perfectly parallel lines, the system has no solution. If you see two perfectly overlapping lines, the system has infinitely many solutions.
Be careful though! Sometimes the SAT will try to trick you by giving you two lines that are very close to being parallel, but actually intersect somewhere far away. They may also do this with lines that look like they overlap.
Practice Question #9
\[20y = x + 1\]
\[41y = 2x\]
How many solutions does the given system of equations have?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is B. We can graph the system by typing its two equations into the expression list. We find that the two lines intersect at a single point, \((-41, -2)\), so the system has exactly one solution.
Answer A is incorrect because these lines do not exactly overlap (graph them and zoom in close to see for yourself!). For example, the point \((0, 0)\) is a solution to the second equation because \(41(0) = 2(0)\), but it is not a solution to the first equation because \(20(0) \neq 1\).
Answer C is incorrect, because a system of linear equations can never intersect in exactly two points. Systems of linear equations will always have \(0\), \(1\), or infinitely many solutions.
Answer D is incorrect. Although these lines may appear to be parallel (in which case the system would have no solution), they actually are not! They intersect at the point \((-41, -2)\).
Not quite...
Show Answer
The correct answer is B. We can graph the two equations by typing them into the expression list. We find that the two lines intersect at a single point, \((-41, -2)\), so the system has exactly one solution.
Answer A is incorrect because these lines do not exactly overlap (graph them and zoom in close to see for yourself!). For example, the point \((0, 0)\) is a solution to the second equation because \(41(0) = 2(0)\), but it is not a solution to the first equation because \(20(0) \neq 1\).
Answer C is incorrect, because a system of linear equations can never intersect in exactly two points. Systems of linear equations will always have \(0\), \(1\), or infinitely many solutions.
Answer D is incorrect. Although these lines may appear to be parallel (in which case the system would have no solution), they actually are not! They intersect at the point \((-41, -2)\).
Number of (Real) Solutions Part 3: Systems of Nonlinear Equations
Sometimes the SAT will give you a system with either one or two nonlinear (usually quadratic) equations. The same rules apply as for systems of linear equations: every point of intersection represents a solution, exactly overlapping graphs represent infinitely many solutions, and two graphs that do not touch represent no solutions.
The most common form of a system of equations involving nonlinear equations is finding the height at which a horizontal line will intersect a parabola at exactly one point. In this situation, the height of the horizontal line will always be the same as the height of the vertex, so you can save some time finding the answer by simply graphing the parabola and clicking on its vertex to find the \(y\)-coordinate. This is the height at which the horizontal line will intersect the parabola at exactly one point.
Practice Question #10
In the \(xy\)-plane, the graph of the equation \(y = -x^2 + 9x - 100\) intersects the line \(y=c\) at exactly one point. What is the value of \(c\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. We can graph the expression \(y = -x^2 + 9x - 100\) and the line \(y = c\) with \(c\) as a slider. If we set \(c\) to each of the four answer choices and zoom out on the graph, we find that \(c = -\frac{319}{4}\) is the only value that makes the line \(y = c\) intersect the parabola at the vertex, exactly one point.
Answer A is incorrect and may result from conceptual or calculation errors. The line \(y = -\frac{481}{4}\) intersects the parabola at two points, not one.
Answer B is incorrect and may result from conceptual or calculation errors. The line \(y = -100\) intersects the parabola at two points, not one.
Answer C is incorrect and may result from conceptual or calculation errors. The line \(y = -\frac{319}{4}\) does not intersect the parabola.
Not quite...
Show Answer
The correct answer is C. We can graph the expression \(y = -x^2 + 9x - 100\) and the line \(y = c\) with \(c\) as a slider. If we set \(c\) to each of the four answer choices and zoom out on the graph, we find that \(c = -\frac{319}{4}\) is the only value that makes the line \(y = c\) intersect the parabola at the vertex, exactly one point.
Answer A is incorrect and may result from conceptual or calculation errors. The line \(y = -\frac{481}{4}\) intersects the parabola at two points, not one.
Answer B is incorrect and may result from conceptual or calculation errors. The line \(y = -100\) intersects the parabola at two points, not one.
Answer C is incorrect and may result from conceptual or calculation errors. The line \(y = -\frac{319}{4}\) does not intersect the parabola.
Graph Features
Identifying features of a graph is perhaps the most well-known usage of Desmos. Try graphing the function \(x^3 - 6x^2 + 11x - 6\) and selecting it either in the expression list or on the graph itself. The gray dots along the graph represent key features, including the following:
- Points of intersection with other graphs
- \(x\)-intercepts (roots)
- \(y\)-intercept
- Local minimums and maximums
- Global minimums and maximums
You can hover over or click the gray dots to see the coordinates of each feature. This simple yet powerful feature is a great way to quickly and accurately identify the features of a graph without having to do any algebraic work, and is especially helpful in questions that ask for the coordinates of a specific feature, such as the vertex of a parabola.
Play around with graphing different types of functions and identifying their key features until you feel comfortable.
Practice Question #11
\[y = x^2 - 9x + 14\]
The given parabola has a two \(x\)-intercepts at the points \((a, 0)\) and \((b, 0)\), a \(y\)-intercept at the point \((0, c)\) and a vertex at the point \((d, e)\). What is the value of \(a + b + c + d + e\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is A. We can graph the given expression and find the coordinates of the \(x\)-intercepts, \(y\)-intercept, and vertex by clicking on the gray dots. The coordinates of the \(x\)-intercepts are \((2, 0)\) and \((7, 0)\), the coordinates of the \(y\)-intercept is \((0, 14)\), and the coordinates of the vertex are \((4.5, -6.25)\). Adding these values together gives us \(2 + 7 + 14 + 4.5 + (-6.25) = 21.25\).
Answer B is incorrect and results from omitting the \(y\)-coordinate of the vertex when taking the sum of values.
Answer C is incorrect and results from omitting the \(y\)-coordinate of the \(y\)-intercept when taking the sum of values.
Answer D is incorrect and results from forgetting both the \(x\) and \(y\)-coordinates of the vertex when taking the sum of values.
Not quite...
Show Answer
The correct answer is A. We can graph the given expression and find the coordinates of the \(x\)-intercepts, \(y\)-intercept, and vertex by clicking on the gray dots. The coordinates of the \(x\)-intercepts are \((2, 0)\) and \((7, 0)\), the coordinates of the \(y\)-intercept is \((0, 14)\), and the coordinates of the vertex are \((4.5, -6.25)\). Adding these values together gives us \(2 + 7 + 14 + 4.5 + (-6.25) = 21.25\).
Answer B is incorrect and results from omitting the \(y\)-coordinate of the vertex when taking the sum of values.
Answer C is incorrect and results from omitting the \(y\)-coordinate of the \(y\)-intercept when taking the sum of values.
Answer D is incorrect and results from forgetting both the \(x\) and \(y\)-coordinates of the vertex when taking the sum of values.
Composite Functions
Sometimes the SAT will ask you to evaluate a composite function, such as \(f(g(x))\). This can be easily modeled with Desmos. Try graphing the functions \(f(x) = 2x + 3\) and \(g(x) = -5x + 2\). If you want to create the composite function \(f(g(x))\), just type \(f(g(x))\) into the expression list and Desmos will automatically evaluate the composite function for you. Try graphing \(f(g(x))\) with the functions \(f(x) = 2x + 3\) and \(g(x) = -5x + 2\) to find the \(y\)-intercept of this new composite function. You can also use Desmos to evaluate the composite function at a specific value of \(x\), just as you would evaluate a standard function. Try evaluating \(f(g(4))\) with the functions \(f(x) = 2x + 3\) and \(g(x) = -5x + 2\). These are all simple and foolproof ways to navigate composite functions on the SAT without ever touching a pencil or paper.
Other times, the question may define a function \(g(x)\) as the function \(f(x)\) with some mathematical operation on the input of \(f(x)\). For example, let's say the question defines the function \(g(x) = f(x+5)\), where \(f(x) = 4x^2 + 64x + 262\). Again, you can type these two expressions into Desmos and Desmos will automatically graph both functions. Try graphing the functions \(f(x) = 4x^2 + 64x + 262\) and \(g(x) = f(x + 5)\), and finding the value of \(x\) for which \(g(x)\) reaches its minimum. You should see both the graph of \(f(x)\) and the graph of \(g(x)\) on the same screen. Since \(g(x)\) is an upward opening parabola, its minimum is at the vertex. You can then find the vertex as we covered in the previous section. You should find that the vertex of \(g(x)\) is at the point \((-13, 6)\), so the minimum value of \(g(x)\) is \(-13\).
Practice Question #12
If \(f(x) = x^2 - 10\) and \(g(x) = \frac{1}{2}x + 5\), what is the value of \(f(g(4)) - g(f(4))\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. We can start by defining \(f(x)\) and \(g(x)\) in Desmos by typing them into the expression list as given. Then we can evaluate \(f(g(4)) + g(f(4))\) by typing it into the expression list, which (Desmos) evaluates as \(31\).
Answer A is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Answer B is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Answer D is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Not quite...
Show Answer
The correct answer is C. We can start by defining \(f(x)\) and \(g(x)\) in Desmos by typing them into the expression list as given. Then we can evaluate \(f(g(4)) + g(f(4))\) by typing it into the expression list, which (Desmos) evaluates as \(31\).
Answer A is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Answer B is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Answer D is incorrect and may result from conceptual or calculation errors. This not value of \(f(g(4)) - g(f(4))\).
Geometry
Triangles
For triangles, it is generally not recommended to use Desmos. Triangle problems often involve unknown side lengths/angles, and Desmos has no quick and easy way to represent this. Instead, we suggest that you brush up on your triangle knowledge by making sure you are familiar with the following concepts:
- Pythagorean Theorem
- Isosceles Triangles
- Similar Triangles
- SOHCAHTOA
Since Desmos cannot be used to easily visualize these problems, we recommend that you sketch them out on scratch paper so that you don't have to do it all in your head.
Circles
Like triangles, geometry problems with circles are often easiest solved on paper rather than graphing. Unless the question explicitly gives you an equation for a circle with \(x\) and \(y\), you should not be using Desmos to solve these problems. Instead, try to familiarize yourself with the following:
- Area
- Circumference
- Radius and diameter
- Arc length
- Radians and degrees (and converting between the two)
To identify the highest and lowest points on a circle, graph the circle and select it either in the expression list or on the graph itself. You can click on the gray dots at the top and bottom of the circle to see the coordinates of the highest and lowest points. From there, you can find the diameter of the circle by calculating the vertical distance from the top and bottom points the circle. This is a great way to quickly find the radius of a circle without having to do any algebraic work.
Practice Question #13
In the \(xy\)-plane, the graph of \(2x^2 - 6x + 2y^2 + 2y = 45\) forms a circle. What is the radius of the circle?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is A. As mentioned before, if the problem gives you an explicit equation with \(x\) and \(y\), Desmos can come in handy! By graphing the equation and clicking on the top and bottom points of the circle, we see that the top of the circle is at \(y = 4.5\) and the bottom is at \(y = -5.5\). We can find the diameter by taking the difference between the two \(y\) values, so the diameter is \(4.5 - (-5.5) = 10\). The radius is half the diameter, so the radius is \(\frac{10}{2} = 5\).
Answer B is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Answer C is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Answer D is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Not quite...
Show Answer
The correct answer is A. As mentioned before, if the problem gives you an explicit equation with \(x\) and \(y\), Desmos can come in handy! By graphing the equation and clicking on the top and bottom points of the circle, we see that the top of the circle is at \(y = 4.5\) and the bottom is at \(y = -5.5\). We can find the diameter by taking the difference between the two \(y\) values, so the diameter is \(4.5 - (-5.5) = 10\). The radius is half the diameter, so the radius is \(\frac{10}{2} = 5\).
Answer B is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Answer C is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Answer D is incorrect and may result from conceptual or calculation errors. This is not the radius of the circle.
Sliders
Whenever a problem contains an unknown free variable (usually denoted by \(a\), \(b\), \(c\), \(k\)), you should probably create a slider. (See Desmos 101 if you forgot how!)
Once you've created a slider to represent the free variable in the problem, you can literally "slide" it around until the graph matches some desired property (whatever the question tells you about the graphs). For example, consider the following question: For what value of \(a\) does the line \(y = ax\) intersect the point \((2,24)\)?
In this case, you can graph the line \(y = ax\) and the point \((2,24)\), and create a slider for \(a\). Note that you may have to change the slider bounds so that you can slide \(a\) to the correct value. Try it for yourself! Graph the line \(y = ax\) and the point \((2,24)\) with a slider for \(a\). Then slide \(a\) until it intersects the point. You should find that the line intersects the point when \(a = 12\).
This method of creating a slider for an unknown free variable can be applied to all sorts of questions, whether it be finding the value of \(a\) that makes a line tangent to a parabola, finding the value of \(a\) that makes two lines parallel, finding the value of \(a\) that makes two expressions equivalent, etc. Bottom line: if you are given a free variable to solve for, your instinct should be to consider making a slider for it.
Practice Question #14
\[y=2x^2 - 21x + 64\]
\[y=3x + c\]
In the given system of equations, \(c\) is a constant. The graphs of the equations in the given system intersect at exactly one point, \((x, y)\), in the \(xy\)-plane. What is the value of \(x\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. We can graph the two equations and create a slider for \(c\). By moving the slider to test different values of \(c\), we find that the line intersects the parabola at exactly one point when the line is tangent to the right side of the curve, specifically when \(c = -8\). We can click on the gray dot at the intersection to find the coordinates of the point of intersection, which is \(6, 10\). So the value of the \(x\)-coordinate when the two graphs intersect at exactly one point is \(6\).
Answer A is incorrect because it is the value of \(c\) when the graphs of the equations in the given system intersect at exactly one point, not the value of the \(x\)-coordinate at which they intersect.
Answer B is incorrect because it is the additive inverse (negative) of the correct value of \(x\).
Answer D is incorrect because it is the additive inverse (negative) of \(c\) when the graphs of the equations in the given system intersect at exactly one point, not the value of the \(x\)-coordinate at which they intersect. Wrong sign and wrong variable.
Not quite...
Show Answer
The correct answer is C. We can graph the two equations and create a slider for \(c\). By moving the slider to test different values of \(c\), we find that the line intersects the parabola at exactly one point when the line is tangent to the right side of the curve, specifically when \(c = -8\). We can click on the gray dot at the intersection to find the coordinates of the point of intersection, which is \(6, 10\). So the value of the \(x\)-coordinate when the two graphs intersect at exactly one point is \(6\).
Answer A is incorrect because it is the value of \(c\) when the graphs of the equations in the given system intersect at exactly one point, not the value of the \(x\)-coordinate at which they intersect.
Answer B is incorrect because it is the additive inverse (negative) of the correct value of \(x\).
Answer D is incorrect because it is the additive inverse (negative) of \(c\) when the graphs of the equations in the given system intersect at exactly one point, not the value of the \(x\)-coordinate at which they intersect. Wrong sign and wrong variable.
Tables & Regressions
In Desmos, a table is a set of \(x\) and \(y\) values representing points in the \(xy\)-plane, usually the inputs and outputs of a function. You can create a table by pressing the gray plus sign in the top left of the calculator and selecting "table." You can also type table into an empty expression line. Note that you cannot create a table using \(x\) and \(y\) as variable names, as these are reserved for the \(x\) and \(y\) axes. Instead, you can use \(x_1\) and \(y_1\), \(a\) and \(f(a)\), or any other non-reserved variable name.
An incredibly useful feature of tables in Desmos is that you can add regressions to them. A regression is a tool that fits a function to a given data set, and you can add a regression to a table by clicking the regression icon in the upper left of any table with at least two points.
Try creating a table for a function \(h(x)\) where \(h(28) = 15\) and \(h(26) = 22\). Use \(x_1\) as the independent/input variable and \(y_1\) as the dependent/output variable in the table header. Then click the regression icon in the upper left corner of the table. You can see the characteristics of the regression in the expression list, including the slope and \(y\)-intercept of the line that best fits the data.
You can also create a custom regression using a tilde (~) in place of an equal sign. Try typing \(y_1 \sim mx_1 + b\) in a new expression line to create a custom linear regression. In this case, both this new custom regression and the regression in the table are the same because we have defined both as linear, but keep in mind that the custom regression allows you to create a regression for any class of function you want.
You can use this method to find the equation or the value of some parameter in the equation of a parabola or any other function that fits a given data set. Remember that the closer the \(R^2\) value is to 1, the better the regression fits the data, so an \(R^2\) value of 1 means the regression fits perfectly.
Practice Question #15
| \(x\) |
\(f(x)\) |
| \(1\) |
\(5\) |
| \(3\) |
\(13\) |
| \(5\) |
\(21\) |
Some values of the linear function \(f\) are shown in the table above. Which of the following defines \(f\)?
Correct! Move on for more Desmos SAT tips, or see how we approached this question.
The correct answer is C. We can create a table in Desmos with the given values of \(x\) and \(f(x)\) using some non-reserved variable names, such as \(x_2\) and \(y_2\) (\(x_1\) and \(y_1\) were used in the example, sorry!). We know that \(f\) is linear, so we can click the regression icon in the top left corner of the table to find the equation of a line that best fits the data. In the regression parameters, we find that the slope is \(4\) and the \(y\)-intercept is \(1\), so the equation of the linear function \(f\) is \(f(x) = 4x + 1\).
Answer A is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
Answer B is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
Answer D is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
Not quite...
Show Answer
The correct answer is C. We can create a table in Desmos with the given values of \(x\) and \(f(x)\) using some non-reserved variable names, such as \(x_2\) and \(y_2\) (\(x_1\) and \(y_1\) were used in the example, sorry!). We know that \(f\) is linear, so we can click the regression icon in the top left corner of the table to find the equation of a line that best fits the data. In the regression parameters, we find that the slope is \(4\) and the \(y\)-intercept is \(1\), so the equation of the linear function \(f\) is \(f(x) = 4x + 1\).
Answer A is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
Answer B is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
Answer D is incorrect may result from conceptual or calculation errors. This is not the equation for the linear function \(f\).
General SAT Advice
Here are some final pieces of wisdom as previous SAT takers:
- There is no substitute for practice! Head over to College Board and take plenty of the free, full-length practice tests on Bluebook to familiarize yourself with the pace and content of the test. Be sure to practice as you would on a real test, and develop a consistent routine so that you feel prepared and comfortable on test day. When you're done, review your practice results to see what you need to work on.
- Although Desmos is incredibly powerful, you may want to familiarize yourself with the following: conversion between standard form, vertex form, and factored form for a quadratic; conversion between standard form, slope-intercept form, and point-slope form for a line; the discriminant; the product and sum of solutions to a quadratic; the equation for a circle; SOHCAHTOA.
- If you've been stuck on a question for a while, consider flagging it in Bluebook and returning to it later. Looking at a question a second time can bring up new ideas, and you don't want to risk omitting answers to easy questions because you got stuck on a single hard question. At worst, there is no penalty for guessing!
- Before test day, make sure you get a good night's sleep and eat a full meal. The last thing you want is to lose points due to fatigue (which is somewhat understandable given the opening reading section).
Congratulations on completing SAT Math! With Desmos by your side, you're ready to tackle the SAT. We hope you found this guide helpful and that it has made your SAT experience a little less stressful. Remember, the SAT is just one part of your college application, and there are many paths to success. Tests and college decisions do not define your worth or potential. Keep working hard, stay positive, and believe in yourself. You've got this!
Feeling adventurous? Ready to take your Desmos skills to the next level? Check out Secret Tips + Artwork to discover exclusive features you won't find anywhere else. Keep exploring and have fun!
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