Desmos on the SAT

The digital SAT has a Desmos graphing calculator built into the testing app, and you can use it on every math question. (The testing app also has a scientific calculator mode, and you can switch between the two during the Math section.) Used well, it turns many “solve this” questions into “graph this and click the point”, and it’s a great way to check an answer you found by hand. This page shows the moves that come up again and again, what to type, and when it’s faster to just do the algebra in your head.

Key ideas

The basics

• Each line you type in the left panel is an expression. Desmos graphs it right away.
• Type ^ for a power, sqrt for a square root, and / for a fraction.
• Click a curve to see grey dots at its important points: x-intercepts (zeros), y-intercept, maximum or minimum, and intersections with other curves. Click a grey dot to see its coordinates.
• If you can’t see a point, the window is the problem. Zoom out (scroll or use the minus button) or use the zoom-fit button.

You can also bring an approved handheld calculator, but Desmos is usually faster for anything involving a graph.

Solving an equation: graph both sides

To solve an equation in one variable, graph each side as its own function and find where the graphs cross:

• type y = left side on one line and y = right side on the next;
• click the intersections. Their x-coordinates are the solutions.

This works for any kind of equation: linear, quadratic, radical, exponential. It’s the same idea as solving equations graphically. For an equation set equal to $0$, one graph is enough: graph $y =$ (the expression) and read the zeros.

Solving a system

Type both equations exactly as given. Desmos is happy with 3x + 2y = 16, so you don’t need to solve for $y$ first. Click the intersection to read the solution $(x, y)$. See solving linear systems by graphing.

• One intersection: one solution.
• Parallel lines: no solution.
• The same line twice: infinitely many solutions.

Quadratics: zeros, vertex, maximum, minimum

Graph $y = ax^2 + bx + c$ and click the parabola. The grey dots show the zeros, the y-intercept, and the vertex (labelled as a maximum or minimum point). That answers questions about the quadratic formula, vertex form, and maximum and minimum values in seconds.

Sliders for unknown constants

If an equation has a letter other than $x$ and $y$ in it, like y = kx + 3, Desmos offers to add a slider for $k$. Drag the slider and watch the graph change. This is perfect for questions like “for what value of $k$ does the system have no solution?” Drag until the lines look parallel, then confirm the exact value with algebra (or by typing the value, like k = -3).

Regression with a table

For a set of data points:

1. Add a table (the + button, then “table”). Enter the x-values in the $x_1$ column and the y-values in the $y_1$ column.
2. On a new line, type y_1 ~ mx_1 + b (the ~ means “fit this model”).
3. Desmos shows $m$, $b$, and $r^2$ (and $r$). For a curve of best fit, change the model, e.g. y_1 ~ ax_1^2 + bx_1 + c or y_1 ~ ab^(x_1).

See linear regression for what these numbers mean.

Statistics with lists

Type a list in square brackets and use the built-in functions:

Type this	You get
mean([4, 7, 7, 10])	the mean, $7$
median([4, 7, 7, 10])	the median, $7$
stdev([4, 7, 7, 10])	the sample standard deviation
stdevp([4, 7, 7, 10])	the population standard deviation
max(...), min(...)	the largest and smallest values

You can name a list, like L = [4, 7, 7, 10], and then type mean(L). The SAT mostly asks about standard deviation by comparing spreads, so you’ll rarely need its exact value. See measures of central tendency and standard deviation.

Checking equivalent expressions

To test whether two expressions are equivalent, graph y = first expression and y = second expression. If the graphs lie exactly on top of each other everywhere, the expressions are equivalent. If you can see two different graphs anywhere, they aren’t.

Evaluating functions

Define the function once, like f(x) = 3x^2 - 5x + 1, then type f(4) on a new line. Desmos shows the value. You can even type f(4) - f(-2).

When not to use Desmos

Desmos isn’t always the fastest tool. Skip it when:

• the algebra is one or two quick steps, like $3x + 5 = 20$;
• the question asks for an expression or an equation (like “which expression is equivalent…?”) where reading the choices is faster than typing all four;
• the question is about meaning, like “what does the $12$ represent in this model?”

A good habit: solve it by hand if you can, then use Desmos as a quick check on the harder questions.

Answer formats

Desmos shows decimals, rounded to a few places. On a multiple-choice question, compare the decimal to the choices (you may need to convert a choice like $\dfrac{-3 + \sqrt{65}}{4}$ to a decimal). On a student-produced response (fill-in) question:

• you can enter a fraction or a decimal, like $7/2$ or $3.5$ (no mixed numbers);
• a positive answer can use up to 5 characters and a negative answer up to 6 (the minus sign counts);
• if a decimal is too long, fill all the spaces: for $\dfrac{2}{3}$, enter $2/3$, $.6666$, $.6667$, $0.666$, or $0.667$, but not $0.67$.

Worked examples

Example 1: Graph both sides

Solve $x^2 - 2x - 3 = x + 1$.

Solution.

By hand. Move everything to one side and factor:

$$\begin{aligned} x^2 - 2x - 3 &= x + 1 \\ x^2 - 3x - 4 &= 0 \\ (x - 4)(x + 1) &= 0 \end{aligned}$$

So $x = 4$ or $x = -1$.

In Desmos. Type y = x^2 - 2x - 3 and y = x + 1. Click where they cross. The grey points are $(-1, 0)$ and $(4, 5)$, so the solutions are the x-coordinates, $x = -1$ and $x = 4$.

The solutions of $x^2 - 2x - 3 = x + 1$ are the x-coordinates of the intersections: $x = -1$ and $x = 4$.

Careful: the answer is the x-coordinate of each point. A common trap is to report $5$ (the y-coordinate) as a solution.

Example 2: A system of linear equations

$$\begin{aligned} 3x + 2y &= 16 \\ x - y &= 2 \end{aligned}$$

If $(x, y)$ is the solution of the system, what is the value of $x + y$?

Solution.

By hand. From the second equation, $x = y + 2$. Substitute into the first:

$$\begin{aligned} 3(y + 2) + 2y &= 16 \\ 5y + 6 &= 16 \\ y &= 2 \end{aligned}$$

Then $x = 2 + 2 = 4$, so $x + y = 6$.

In Desmos. Type 3x + 2y = 16 and x - y = 2 exactly as written. Click the intersection: $(4, 2)$. So $x + y = 4 + 2 = 6$.

Check: $3(4) + 2(2) = 16$ ✓ and $4 - 2 = 2$ ✓.

Example 3: A slider for “no solution”

In the system below, $k$ is a constant. For what value of $k$ does the system have no solution?

$$\begin{aligned} 6x - ky &= 5 \\ 4x + 2y &= 9 \end{aligned}$$

Solution.

By hand. A linear system has no solution when the lines are parallel but different. Parallel lines have the same ratio of $x$-coefficient to $y$-coefficient:

$$\frac{6}{4} = \frac{-k}{2} \quad\Rightarrow\quad 12 = -4k \quad\Rightarrow\quad k = -3$$

Check the lines are different: with $k = -3$ the first equation is $6x + 3y = 5$, or $2x + y = \dfrac{5}{3}$. The second is $2x + y = \dfrac{9}{2}$. Same left side, different right side, so they’re parallel and never meet. ✓

In Desmos. Type 6x - ky = 5 and click “add slider” for $k$. Type 4x + 2y = 9 on the next line. Drag $k$ until the two lines are parallel. It happens at $k = -3$ (type k = -3 to land on it exactly). For any other $k$, the lines cross somewhere, even if it’s off the screen.

Example 4: Regression from a table

A student measures the height of a seedling each week.

Week, $x$	1	2	3	4	5
Height (cm), $y$	3.1	4.9	7.2	8.8	11.0

Find the line of best fit, and use it to predict the height in week $8$, to the nearest tenth of a centimetre.

Solution.

In Desmos. Add a table, enter the weeks in $x_1$ and the heights in $y_1$. Then type y_1 ~ mx_1 + b. Desmos shows

$$m = 1.97, \qquad b = 1.09, \qquad r^2 \approx 0.9977$$

So the line of best fit is $y = 1.97x + 1.09$. The value of $r^2$ is very close to $1$, so the line fits the data very well.

Predict. Substitute $x = 8$:

$$y = 1.97(8) + 1.09 = 15.76 + 1.09 = 16.85$$

The model predicts a height of about $16.9$ cm in week $8$. (You can also type 1.97(8) + 1.09 straight into Desmos.)

Making sense of it. The slope $1.97$ means the seedling grows about $2$ cm per week. Predicting week $8$ goes a little beyond the data, so it’s a reasonable estimate, but the further you go beyond the data, the less you should trust the model.

Common mistakes

Reading the wrong coordinate. When you graph both sides of an equation in $x$, the solutions are the x-coordinates of the intersections. When you graph a system, the answer is the whole point $(x, y)$. Reread what the question asks for, like $x$, $y$, or $x + y$.

Missing a point outside the window. If the graphs don’t seem to cross, or you only see one of two intersections, zoom out before deciding. Lines that look parallel on a small window might cross far away.

Copying a rounded decimal as exact. Desmos might show $(1.266, 0)$ when the exact zero is $\dfrac{-3 + \sqrt{65}}{4}$. On a fill-in question the decimal is fine if it fills the spaces. On multiple choice, convert the choices to decimals and compare, rather than assuming $1.266$ is “the” answer.

Using the wrong angle mode. On the SAT, the built-in Desmos starts in degrees, but the regular desmos.com calculator you practise on starts in radians. Before any trig question, check the angle mode in the settings (the wrench icon), and switch to radians if the question uses radians, or your answers will be way off.

Trusting a slider by eye. Sliders get you close, but “the lines look parallel” isn’t proof. Type the exact value (like k = -3) or confirm with algebra, especially for fill-in questions.

Spending too long typing. If you can solve it in your head in ten seconds, do that. Save Desmos for messy equations, systems, quadratics, data, and checks.

Practice

1. (Warm-up) Solve $5x - 7 = 2x + 11$. What would you type into Desmos to check?

Solution

$$5x - 7 = 2x + 11 \quad\Rightarrow\quad 3x = 18 \quad\Rightarrow\quad x = 6$$

In Desmos, type y = 5x - 7 and y = 2x + 11. They cross at $(6, 23)$, so $x = 6$.

2. (Warm-up) What is the minimum value of $y = x^2 - 6x + 5$?

• A) $-6$
• B) $-4$
• C) $3$
• D) $5$

Solution

B. Graph y = x^2 - 6x + 5 and click the parabola: the minimum point is $(3, -4)$, so the minimum value is $-4$.

By hand: the vertex is at $x = -\dfrac{b}{2a} = \dfrac{6}{2} = 3$, and $y = 9 - 18 + 5 = -4$. (Choice C is the x-coordinate of the vertex, not the minimum value.)

3. (Warm-up) Find the mean and the median of the data set $12, 15, 15, 18, 20, 22, 31$.

Solution

Type L = [12, 15, 15, 18, 20, 22, 31], then mean(L) and median(L).

Mean: $\dfrac{12 + 15 + 15 + 18 + 20 + 22 + 31}{7} = \dfrac{133}{7} = 19$.

Median: the data are already in order and there are $7$ values, so the median is the 4th value, $18$.

4. (Core) What is the positive solution of $2x^2 + 3x - 7 = 0$? (Student-produced response.)

Solution

Graph y = 2x^2 + 3x - 7 and click the positive x-intercept: about $(1.266, 0)$.

Exact value, by the quadratic formula:

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)} = \frac{-3 \pm \sqrt{65}}{4}$$

The positive solution is $\dfrac{-3 + \sqrt{65}}{4} \approx 1.2656$. Acceptable entries: $1.265$ or $1.266$.

5. (Core) Which expression is equivalent to $(x + 3)^2 - (x - 1)^2$?

• A) $8x + 8$
• B) $4x + 8$
• C) $10$
• D) $2x^2 + 4x + 10$

Solution

A. Expand:

$$(x^2 + 6x + 9) - (x^2 - 2x + 1) = 8x + 8$$

Desmos check: graph y = (x + 3)^2 - (x - 1)^2 and y = 8x + 8. The two graphs are the same line.

6. (Core) In the equation $x^2 - 8x + c = 0$, $c$ is a constant. For what value of $c$ does the equation have exactly one real solution?

Solution

In Desmos, graph y = x^2 - 8x + c with a slider for $c$. Drag until the parabola just touches the x-axis: $c = 16$.

By hand: exactly one real solution means the discriminant is $0$:

$$(-8)^2 - 4(1)(c) = 0 \quad\Rightarrow\quad 64 = 4c \quad\Rightarrow\quad c = 16$$

Check: $x^2 - 8x + 16 = (x - 4)^2$, which is $0$ only when $x = 4$. ✓

7. (Core) Let $f(x) = 3x^2 - 5x + 1$. What is the value of $f(4) - f(-2)$?

Solution

Type f(x) = 3x^2 - 5x + 1, then f(4) - f(-2). Desmos shows $6$.

By hand: $f(4) = 48 - 20 + 1 = 29$ and $f(-2) = 12 + 10 + 1 = 23$, so $f(4) - f(-2) = 29 - 23 = 6$.

8. (Challenge) The system below has two solutions, $(x_1, y_1)$ and $(x_2, y_2)$. What is the value of $y_1 + y_2$?

$$\begin{aligned} y &= x^2 - 4x + 1 \\ y &= 2x - 7 \end{aligned}$$

Solution

Graph both equations and click the two intersections: $(2, -3)$ and $(4, 1)$. So $y_1 + y_2 = -3 + 1 = -2$.

By hand: set the right sides equal.

$$\begin{aligned} x^2 - 4x + 1 &= 2x - 7 \\ x^2 - 6x + 8 &= 0 \\ (x - 2)(x - 4) &= 0 \end{aligned}$$

So $x = 2$ or $x = 4$. Then $y = 2(2) - 7 = -3$ and $y = 2(4) - 7 = 1$, and $y_1 + y_2 = -2$.

9. (Challenge) In the system below, $k$ is a constant. For what value of $k$ does the system have infinitely many solutions?

$$\begin{aligned} kx + 3y &= 6 \\ 8x + 12y &= 24 \end{aligned}$$

Solution

Infinitely many solutions means the two equations describe the same line. Divide the second equation by $4$:

$$2x + 3y = 6$$

This matches $kx + 3y = 6$ exactly when $k = 2$.

In Desmos, type kx + 3y = 6 with a slider and 8x + 12y = 24. At $k = 2$ the two lines sit exactly on top of each other. (For every other $k$ they cross at one point, so there’s one solution.)