Sliders for unknown constants

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Use Desmos sliders to solve SAT problems with unknown constants, match graph conditions, test exact values, and apply restrictions.

Difficulty: Intermediate
Estimated time: 35 minutes
Domains: Algebra Advanced Math

What you’ll learn

Spot an unknown constant that a Desmos slider can control.
Make one slider and set its bounds or step size.
Move a graph until it passes through a given point or meets another graph at a given coordinate.
Type exact values instead of trusting how the graph looks.
Use sign, range, and integer restrictions to pick the right value.
Find what the question actually asks for once you know the constant.
Tell when algebra is shorter or more exact.

Prerequisites

Why this matters on the SAT

Find a constant without rearranging the formula

Lots of SAT questions include a letter like $a$ or $k$ and call it a constant. It stands for one fixed number, but you don't know which one yet. It decides where a graph sits or where two graphs cross. A Desmos slider lets you turn that number up and down and watch the graph move, with the rest of the problem still in view.

SAT example

The function $f$ is defined by

f(x)=a(x-4)^2-3,

where $a$ is a constant. The graph of $f$ passes through the point $(2,9)$ . What is the value of $f(0)$ ?

Fast Desmos solution

Type f(x)=a(x-4)^2-3, add a slider for $a$ , and plot $(2,9)$ . At $a=3$ , the curve passes right through the point. Now type f(0). Desmos shows $45$ , so the answer is C.

Notice that $3$ is sitting there as choice A. That's the trap in questions like this one. The slider finds the constant, but the question asks for something else, here $f(0)$ . So once you have the constant, read the question again and give it what it asks for: find the constant, then answer the question.

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With a = 3, the curve passes through (2, 9), and f(0) = 45.

When should you use a slider?

A slider shines when one unknown number moves a graph and you can see the goal. When a line of algebra gets you there first, use the algebra.

Start with a slider when…

One unknown constant changes the graph, like $a$ in $f(x)=a(x-4)^2-3$ .
You can see the target, like a point the graph has to pass through or an $x$ -coordinate where two lines have to cross.
The answer choices or a restriction leave only a few sensible values to try, like four choices for $c$ .
You’re not sure what the constant does, and moving it back and forth would show you.
The Desmos setup is short and easier to follow than the algebra, like a constant tucked inside an absolute value.

Use algebra when…

One substitution gives you the constant right away. If $f(x)=kx^2$ and $f(2)=12$ , then $4k=12$ , so $k=3$ .
The exact answer is an awkward fraction, like $\frac{7}{3}$ , which is hard to land on by dragging a slider.
The problem has two or more unknown constants, like $a$ and $b$ in $f(x)=ax^2+b$ , plus a condition for each. One slider can’t handle that, so it’s a job for the regression workflow in a later lesson.

You’ll spot these questions by wording like “where $k$ is a constant,” “the graph passes through the point,” “the solution has an $x$ -coordinate of,” “which value of $c$ could make the statement true?”, or “ $a$ is a positive integer.” A question about how many solutions there are, like exactly zero, one, or many, is a different job, and it gets its own solution-count workflow in a later lesson.

Check your understanding:

Here are two problems. In the first, the line $y=3x+b$ passes through $(2,8)$ . In the second, you need the value of $c$ from $1$ , $3$ , $5$ , or $7$ that makes $g(x)=|x-c|+\frac{x^2}{4}$ pass through $(6,12)$ . Which one would you hand to a slider, and why?

Create one slider and see what changes

In Desmos, lowercase $x$ and $y$ are the graphing variables. Any other letter you haven't defined, like $a$ , $b$ , $c$ , or $k$ , can stand for a constant.

Type

y=2x+c.

Desmos offers to add a slider for $c$ . You can also make one yourself by typing c=0 on its own line.

A few settings make the slider much more useful:

If the problem gives a range, set the minimum and maximum to match it.
If the constant has to be an integer, use a step of $1$ .
To explore simple decimals, use a smaller step.
When you want to try an exact value, type it straight into the slider.

To find these settings, click the slider's limits control beside the expression. It opens the minimum, maximum, and Step fields.

Try it yourself:

Set the minimum to $-4$ , the maximum to $5$ , and the step to $1$ . Then move $c$ through its values and watch the $y$ -intercept. What stays the same?

Now the slider stops only on the integers from $-4$ to $5$ . The line slides up and down, and the $y$ -intercept changes with it, but the slope stays $2$ the whole time.

Check your understanding:

Why is a step of $1$ better than a step of $0.1$ when $c$ has to be an integer? And at $c=-4$ , what are the line’s slope and $y$ -intercept?

Common mistake:

Watch for two slips. The first is using $x$ as the constant. Desmos treats $x$ as the horizontal variable, so typing x=0 draws a vertical line instead of making a slider. The second is defining the same constant on two lines, like c=0 and c=3. Desmos can’t tell which value you mean and shows an error. Pick a letter other than $x$ and $y$ , and define it only once.

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Changing c moves the line up and down. The slope stays 2.

Match the condition, then finish the question

Here's the fact that makes sliders work: a point $(u,v)$ is on a function's graph exactly when the function gives $v$ for the input $u$ . So "the graph passes through $(2,23)$ " means the function gives $23$ when $x=2$ .

Suppose

p(x)=b(x+1)^2-4

passes through $(2,23)$ . What is $p(0)$ ?

Type the function, add a slider for $b$ , and plot the point. Then work through these steps:

Keep the function and the point on screen.
Drag the slider to see which way the curve moves as $b$ changes.
Type exact values into the slider.
Check the value that works in the point condition.
Type the expression the question asks for.

Why type values instead of dragging? Dragging gets you close, and close can look exact on the screen. A typed value, checked in the point condition, settles it.

Check your understanding:

Use the graph and the point. What are the values of $b$ and $p(0)$ ?

Common mistake:

Submitting the slider value. The question might want an output like $p(0)$ , a coordinate, or an expression like $2k$ . Read the last sentence of the question once more before you answer.

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Change b until the curve passes through (2, 23), then find p(0).

Example: Match a system to a stated coordinate

A slider can also move one line in a system. Keep both graphs on screen and change the constant until the lines cross at the spot the question describes.

Worked example

The graphs of

y=2x+5

and

y=kx-1,

where $k$ is a constant, intersect at a point with an $x$ -coordinate of $3$ . What is the value of $2k$ ?

A
$4$
B
$6$
C
$8$
D
$10$

Step 1

Find the target

The two lines have to cross where $x=3$ . The question doesn't give the $y$ -coordinate, and you don't need it: the intersection will show you.

Step 2

Graph both lines

Type both equations and add a slider for $k$ . Zoom out far enough to see where the lines cross.

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Change k until the lines cross at x = 3.

Step 3

Set the constant and check it

At $k=4$ , the lines cross at $(3,11)$ . Check that point in both equations:

2(3)+5=11

and

4(3)-1=11.

The point works in both, so $k=4$ is exact.

Step 4

Answer the question

The question asks for $2k$ , not $k$ :

2k=2(4)=8.

The answer is C.

Common mistake: Picking A. That's $k$ itself, one step short of $2k$ .

Apply every restriction

Sometimes more than one slider value makes the graph fit. When that happens, a restriction in the problem, like a sign or a range, tells you which one counts.

Take

g(x)=|x-c|+2,

where $c>0$ . The graph passes through $(1,7)$ . What is the value of $c$ ?

The point gives

|1-c|+2=7,

so $|1-c|=5$ . That means $1-c=5$ or $1-c=-5$ , so $c=-4$ or $c=6$ . Both graphs pass through $(1,7)$ . The restriction $c>0$ leaves

c=6.

Whenever a constant has to be positive, negative, an integer, or inside an interval, build that into the slider's bounds, or keep the restriction written next to the graph. Here, a minimum of $0$ means dragging never even reaches $c=-4$ . A value that fits the graph but breaks a rule in the problem is never the answer.

Check your understanding:

Why isn’t $c=-4$ an answer, even though its graph passes through $(1,7)$ ?

Common mistake:

Stopping at the first slider value that works. It’s tempting the moment the graph hits the point, but another value may fit too, and then the restriction decides.

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Both c = -4 and c = 6 hit the point, but only one of them fits c > 0.

Finish the solution

Now you finish one. Watch the range the problem gives for $d$ : it tells you how to set up the slider.

Finish the solution

The function $q$ is defined by

q(x)=d(x-1)^2+2,

where $1<d<4$ . The graph of $q$ passes through the point $(4,24.5)$ . What is the value of $q(-1)$ ?

First steps

The function, the point, and a slider for $d$ are already in the calculator.
Set the slider's minimum to $1.5$ , maximum to $3.5$ , and step to $0.5$ . Those are the half-steps that fit inside $1<d<4$ .
Type values until the curve passes through $(4,24.5)$ .
Check that $q(4)=24.5$ .
Type q(-1) and submit that value.

Finish it

Your answer

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Set the range and a 0.5 step, match (4, 24.5), then find q(-1).

Practice problems

Your turn. For each problem, decide first whether a slider or a quick substitution gets you there faster.

Practice problem

The function $f$ is defined by

f(x)=mx-4,

where $m$ is a constant. The graph of $f$ passes through $(5,11)$ . What is the value of $f(2)$ ?

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Practice problem

In the system below, $a$ is a constant:

\begin{aligned} 2x-3y&=1\\[1.4em] ax+6y&=2a. \end{aligned}

The system has a solution with $x=1$ . What is the value of $a$ ?

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Practice problem

The function $g$ is defined by

g(x)=(x-c)^2+2,

where $c>1$ . The graph of $g$ passes through $(1,27)$ . What is the value of $c+2$ ?

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Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Use a letter other than $x$ and $y$ for the constant, and define it only once.
Drag the slider to see what the constant does, then type exact values.
Match every point or intersection the question gives you.
Set the slider's bounds and step to fit any sign, range, or integer restriction.
Check the final value in the original equation or function.
Find the constant, then answer the question: an output, a coordinate, or an expression.
Use algebra when one substitution finds the constant or the answer is an awkward fraction.

Next lesson

How many solutions?

Use intersections, tangency, and graph behavior to tell when a parameter creates zero, one, or multiple solutions.

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Practice

Practice this lesson

595 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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