Custom regression for one unknown parameter

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Learn to use one fact about a function and a custom Desmos regression to find one unknown constant in a linear or nonlinear function model.

Difficulty: Advanced
Estimated time: 40 minutes
Domains: Algebra Advanced Math

What you’ll learn

Find one unknown constant in a function from one fact about it, like $f(4)=16$ .
Decide quickly whether to plug in by hand or let Desmos do the algebra.
Type the fact into Desmos with a tilde, $\sim$ , keeping the function exactly as given.
Use the value Desmos finds to answer the real question, then check it.

Prerequisites

Why this matters on the SAT

Let Desmos find the missing constant

Some SAT questions give you a function with one missing number, plus one fact about the function, like $f(4)=16$ . Find the missing number, and the rest of the question falls into place.

That missing number is called a parameter: a constant that shapes the function, like its slope, stretch or shift. You could solve for it by hand. Or you could type the fact into Desmos with a tilde, $\sim$ , where the equals sign would go, and let Desmos find the value that makes it true. Typing a fact this way is called custom regression, or a regression for short. You give Desmos your exact function and one fact, and it finds the missing number.

Here it is on a typical SAT question.

SAT example

The linear function $f$ is defined by

f(x)=(p-3)x+(2p+4),

where $p$ is a constant. If $f(4)=16$ , what is the value of $f(9)$ ?

Your answer

Fast Desmos solution

Put $4$ in for $x$ , put $16$ on the other side, and use a tilde where the equals sign would go:

16~(p-3)*4+(2p+4)

Desmos finds $p=4$ . Now type the function, f(x)=(p-3)x+(2p+4), and then f(9). Desmos fills in $p=4$ for you and shows $\boxed{21}$ .

Why not solve it by hand? Plugging in gives $4(p-3)+(2p+4)=16$ , with $p$ in two places. You'd distribute, combine like terms and then solve, and every step is a chance to drop a sign. With the tilde, you type the function exactly as it's printed and skip all of that.

Swap $=$ for $\sim$ , and Desmos does the algebra. It isn't always the fastest move, though. Sometimes plugging in finishes the job in one step.

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The first line finds p. The function then uses that value, so f(9) shows 21.

When should you use this method?

Look for questions that give you the function's form, exactly one unknown constant and one fact about the function, like $f(6)=11$ or a point on its graph. The constant shouldn't be squared, under a root, in a denominator or in an exponent.

Before you open Desmos, plug the fact in by hand and look at what's left. It takes a few seconds, and it tells you which way to go. Use a regression only when typing the function as given is faster or safer than finishing by hand.

More than one step? Use a regression when…

the constant appears more than once, like $p$ in $(p-3)x+(2p+4)$
the function is built from another one, like $f(2x+1)+4$ , and rewriting it by hand invites mistakes
the fact comes as a point, graph or table that's quicker to type than to solve

Skip the regression when…

One step finishes it, like $3k=12$ . Solve it by hand, even if the function looks complicated.
You need to see it on the graph, like a line that touches a curve at exactly one point. That's a job for a slider. Confirm the value exactly afterward.

One unknown needs one fact. With two or more unknown constants, one point usually can't pin them all down.

Check your understanding:

The function $f(x)=3x+k$ satisfies $f(5)=19$ . Would you start by plugging in, with a regression or with a slider?

Check your understanding:

A function has the form $f(x)=ax^2+bx+c$ , where $a$ , $b$ and $c$ are all unknown, and you know only one point on its graph. Can one regression pin down the function?

Turn the fact into a regression

When a regression is the right call, writing one takes three moves: put the known input into the function, put the known output on the other side, and use $\sim$ in place of $=$ .

The tilde is what makes this work. An equals sign tells Desmos to graph something or define it. It doesn't ask Desmos to find the unknown.

Here's how common facts translate:

$f(c)=d$ , or a point $(c,d)$ on the graph: type d~ and then the function with $c$ in place of $x$ .
An $x$ -intercept at $x=c$ : the input is $c$ and the output is $0$ .
A $y$ -intercept of $d$ : the input is $0$ and the output is $d$ .
A maximum or minimum at $(c,d)$ : the input is $c$ and the output is $d$ . But a regression only puts $(c,d)$ on the graph. It can't make that point the peak or the valley, so check that part yourself.

Say

q(x)=a(x-2)^2-5

and the graph passes through $(6,11)$ . Type:

11~a(6-2)^2-5

Desmos finds $a=1$ .

One catch: the fact has to tell you something new about $a$ . Try the point $(2,-5)$ instead. Plugging in $x=2$ gives $a(2-2)^2-5$ , which is $-5$ no matter what $a$ is. Every parabola in this family passes through $(2,-5)$ , so that point can't tell them apart. The point $(6,11)$ can.

Common mistake:

Swapping the input and the output. For the point $(6,11)$ , the $6$ goes inside the function and the $11$ goes on the other side: 11~a(6-2)^2-5, not 6~a(11-2)^2-5.

Check your understanding:

The graph of $r$ has an $x$ -intercept at $x=-3$ . How would you write this fact as a regression, or as a table row?

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The fact q(6) = 11 finds a, and q(0) then uses that a.

Use a table when it's clearer

A table holds the same fact as a point. Put the input in x_1 and the output in y_1. Then write the function with three swaps:

x_1 in place of the input variable;
y_1 in place of the output;
$\sim$ in place of $=$ .

For

q(x)=a(x-2)^2-5,

that's:

y_1~a(x_1-2)^2-5

One point and one unknown, so Desmos finds the same $a=1$ .

Try it yourself:

Change the table output from $11$ to $27$ . Predict how $a$ and $q(0)$ will change, then check in the calculator. Reset the example when you're done.

With a table in front of you, you might reach for Desmos's built-in quadratic regression:

y_1\sim ax_1^2+bx_1+c.

It's a natural move, but it won't work: that form has three unknowns, and you have one fact. The problem already told you the function's exact form, so use it.

Check your understanding:

The graph of $r(x)=\frac{c}{x+1}+2c$ passes through $(3,21)$ . Which method should you lead with? Write the entry and say what value Desmos should find.

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The table version keeps the vertex at (2, -5) and finds only a.

Give Desmos only one letter to find

This check takes a second and saves a lot of confusion. Before you run a regression, make sure the only letter without a value is the one you want Desmos to find.

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

$x$ is the input, $a$ is the unknown, and $3$ and $-2$ are fixed numbers. In a table version, x_1 and y_1 hold the known inputs and outputs, and $a$ gets no value, so Desmos is free to find it.

Here's how that goes wrong. Say you made a slider a=2 earlier while exploring. Desmos now treats $a$ as fixed, so the regression won't find it. Delete that line or switch to a fresh letter.

Don't worry about subscripts on the unknown: a and a_1 both work, as long as you use the same name everywhere. The names that must match exactly are the table's. If the columns are labeled x_1 and y_1, your regression has to say x_1 and y_1 too.

Example: Find the constant in a composed model

This one has two functions, one built from the other, and that's where it's easy to slip. Go one step at a time, and keep asking which function the point belongs to.

Worked example

The functions $f$ and

$g$ are defined by

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

and

g(x)=f(2x+1)+4,

where $a$ is a constant. The graph of $g$ passes through $(2,22)$

. What is the value of $f(0)$ ?

Step 1

Count unknowns and facts

There's one unknown, $a$ , and one fact, the point $(2,22)$ . One unknown, one fact, so one regression can find $a$ . The tricky part is which function that point belongs to.

Step 2

Build g from f

The point is on the graph of $g$ , not $f$ , so the regression has to describe $g$ . Put $2x+1$ in place of $x$ in $f$ , then add $4$ :

g(x)=a\big((2x+1)-3\big)^2-2+4.

There's no need to simplify. Desmos handles the nested version fine.

Step 3

Let Desmos find a

Put in the point: $2$ for $x$ , and $22$ on the other side.

22~a((2*2+1)-3)^2-2+4

Desmos finds $a=5$ .

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The point is on g, so the regression uses all of g.

Step 4

Answer the question asked

Careful here: the question asks for $f(0)$ , not $a$ and not $g(0)$ . Go back to $f$ with $a=5$ :

\begin{aligned} f(0) &=5(0-3)^2-2\\[1.4em] &=5(9)-2\\[1.4em] &=\boxed{43}. \end{aligned}

Now check $a=5$ against the fact you were given. Inside $g$ , $(2x+1)-3=2x-2$ and $-2+4=2$ , so

g(2)=5(2\cdot2-2)^2+2=5(4)+2=22.

It matches, so the answer is $43$ .

The algebra was never the hard part here. The hard part was seeing that the point belongs to $g$ . Once you had that, one regression did the rest.

Common mistake:

Plugging $(2,22)$ into $f$ instead of $g$ . That gives $a-2=22$ , so $a=24$ and $f(0)=214$ , nowhere near $43$ . The point is on $g$ , so the regression needs all of $g$ : the $2x+1$ inside and the $+4$ outside.

Use the value, then check it

Desmos remembers the value it finds. So after it finds $a$ , type the original function with $a$ in it, and Desmos fills in the value for you. You never have to retype a rounded decimal.

When the unknown is only multiplied or divided by known numbers or expressions, or added, like the $a$ in $a\cdot3^x$ or $\frac{a}{x-2}$ , Desmos solves for it directly, with no guessing. Those are the questions this lesson covers, and the value is still worth checking against the fact. If the unknown is squared, under a root, in a denominator like $\frac{1}{x-a}$ , or in an exponent like $3^{ax}$ , Desmos has to search for it, and the search can miss. That's beyond this lesson. If you try it, plug the result back in, and a later lesson covers what can go wrong.

What Desmos shows is a decimal, and it may be rounded, so the digits alone can't prove an exact fraction. When a decimal looks like a familiar fraction, the answer choices or a quick conversion can point you to it, the way $9.333$ pointed to $\frac{28}{3}$ for $r$ . Then plug the fraction into the original fact to confirm it.

Last, before you answer, take a few seconds for four checks. Each one catches a common slip:

Does the value make the given fact true?
Did you use the right function, including any shift or stretch? In the example, the point was on $g$ , not $f$ .
Are you answering what's asked? There, it was $f(0)$ , not $a$ .
Did you avoid inputs $x$ can't take, like one that makes a denominator $0$ ?

Finish the solution

The regression and the function $h$ are already in the calculator, and Desmos has found $k$ . Your job is the last step: use $k$ to find the value the question asks for.

Finish the solution

The function $h$ is defined by

h(x)=\frac{k}{x^2-1}+2k,

where $k$ is a constant and $x\ne-1,1$ . If $h(2)=21$ , what is the value of

h\left(\frac12\right)?

First steps

Put $2$ in for $x$ and $21$ on the other side.
Type it with a tilde: 21~k/(2^2-1)+2k.
Desmos finds $k=9$ .

Finish it

Your answer

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Desmos has found k. Add h(1/2) on a new line to finish.

Practice problems

These four don't all want Desmos. Plug each fact in first. If one step finishes it, solve by hand; otherwise use a regression when it's faster or safer.

Practice problem

The function $f$ is defined by

f(x)=a\cdot3^{x-1}-5,

where $a$ is a constant. If $f(1)=4$ , what is the value of $f(0)$ ?

Your answer

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Plugging in by hand wins here. Use the calculator only if you want to check

f(0)

Practice problem

The linear function $g$ is defined by

g(x)=(p-2)x+(3p-1),

where $p$ is a constant. If $g(4)=19$ , what is the value of $g(9)$ ?

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Type the fact with a tilde, keep both p terms, and then evaluate g(9).

Practice problem

The quadratic function $q$ is defined by

q(x)=a(x+1)(x-7),

where $a$ is a constant. The maximum value of $q$ is $8$ , and it occurs at $x=3$ . What is the value of $q(0)$ ?

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Plug in the maximum point

(3,8)

by hand. Use the calculator only if you want to check

q(0)

Practice problem

The functions $f$ and $g$ are defined by

f(x)=\frac{a}{x-2}+1

and

g(x)=2f(x+1)-3,

where $a$ is a constant. If $g(4)=7$ , what is the value of $f(-1)$ ?

Your answer

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Write out all of g, find a from g(4) = 7, then evaluate f(-1).

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Spot it: a function with one unknown constant, plus one fact that pins it down.
Choose: plug in first. One step? Finish by hand. More than that? Use a regression when it saves time or prevents mistakes.
Type it: enter the function exactly as given, with $\sim$ in place of $=$ , and make sure the unknown is the only letter without a value.
Finish: let Desmos reuse the value, answer the question that's asked, and check it against the original fact and any limits on $x$ .

Next lesson

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Combine several exact conditions in one matched-list or table-based custom regression.

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