Hidden functions inside quotients

Your feedback matters:

Lesson progressPractice problems 0/5

Learn to unwrap function values from a quotient, fit the complete nested model in Desmos, and avoid reading an intercept or output from the wrong function.

Difficulty: GOAT Mode
Estimated time: 55 minutes
Domains: Advanced Math

What you’ll learn

Tell which function the table gives and which one the question asks about.
Undo the quotient row by row to find the hidden function's values.
Fit the whole quotient at once with a custom regression.
Keep any known terms around the hidden function in place.
Spot when one table row already answers the question, and take that shortcut.
Read an intercept, value, slope, or vertex from the right function.
Check the function you found in the original quotient.

Prerequisites

Why this matters on the SAT

Recover the function the question actually asks about

Some hard SAT questions give you a table for one function, build that function out of a second one, and then ask about the second one. The table shows you only the outside. The answer is hidden one layer in.

SAT example

The function $g$ is defined by

g(x)=\frac{f(x)}{x+6},

where $f$ is a linear function. The table shows three values of $x$ and the value of $g(x)$ for each.

$x$	$g(x)$
$-20$	$6$
$1$	$-3$
$8$	$0$

Which point is the $y$ -intercept of the graph of $y=f(x)$ in the $xy$ -plane?

Fast Desmos solution

The table gives values of $g$ , but the question asks about $f$ . Since $f$ is linear, write it as

f(x)=mx+b.

Type the table into x_1 and y_1, then fit the whole quotient at once. Use the tilde ~, which tells Desmos to fit, not a minus sign:

y_1~(mx_1+b)/(x_1+6)

Desmos gives $m=3$ and $b=-24$ , so

f(x)=3x-24,

and the $y$ -intercept of $y=f(x)$ is

\boxed{(0,-24)}.

The answer is A.

Here's the trap. The curve on the screen is the quotient $g$ , not $f$ . If you click where it crosses the $y$ -axis, you get

g(0)=\frac{f(0)}{6}=\frac{-24}{6}=-4.

So $(0,-4)$ is a point on $g$ . It's the tempting wrong answer, B.

Calculator loads as you approach

The curve is g and the line is f. Change the 0 in f(0) to try another input.

Read the layers before you open Desmos

A hidden function is a function you never see values of directly. Its values sit inside the definition of another function.

g(x)=\frac{f(x)}{x+6},

here's what each part of the opening problem tells you:

Part of the question	What it tells you
The table	Its numbers are values of $g$
The definition	Each value of $g$ is $f(x)$ divided by the known denominator $x+6$
The function type	Since $f$ is linear, you can write $f(x)=mx+b$
The question	The $y$ -intercept belongs to $f$ , so you need $f(0)$ or $b$
The restriction	The quotient that defines $g$ is undefined at $x=-6$

Before you type anything, finish this sentence:

The table belongs to ______, but the question asks for ______ of ______.

For the opening problem, it reads:

The table belongs to $g$ , but the question asks for the $y$ -intercept of $f$ .

It takes a few seconds, and it heads off the most common mistake in these problems: answering about the wrong function, like choice B here.

Check your understanding:

A table lists values of $h(x)$ , where $h(x)=p(x)/(x-4)$ and $p$ is quadratic. The question asks for the minimum value of $p$ . Which function do you fit, and which function has the minimum you want?

Unwrap the quotient row by row

Multiply by the denominator

g(x)=\frac{f(x)}{x+c},

multiply both sides by $x+c$ . For every input where the denominator isn't zero,

f(x)=(x+c)g(x).

That's not a new model. It's the same relationship, solved for the hidden value.

Try it on the opening table:

\begin{aligned} f(-20)&=(-20+6)(6)=-84,\\[1.4em] f(1)&=(1+6)(-3)=-21,\\[1.4em] f(8)&=(8+6)(0)=0. \end{aligned}

So the table for $g$ hands you three points on $f$ :

(-20,-84),\quad(1,-21),\quad(8,0).

From $(1,-21)$ to $(8,0)$ , the line rises $21$ over a run of $7$ , a slope of $3$ . All three points lie on

f(x)=3x-24.

Let a calculated column do every row

With the inputs in x_1 and the values of $g$ in y_1, add this as a calculated column:

(x_1+6)y_1

The column shows the matching values of $f$ . The nice part is that you can look at the hidden points before you fit anything.

Common mistake:

Multiplying the input by the denominator instead of the output. The relationship is $f(x)=(x+6)g(x)$ , so multiply each value of $g$ , the number in y_1, by its denominator.

Check your understanding:

If $q(x)=r(x)/(2x-1)$ and a table row gives $x=3$ and $q(3)=4$ , what is $r(3)$ ?

Calculator loads as you approach

Change one value of g and watch the f column update. Then reset.

Fit the full quotient directly

Keep the whole quotient in the regression

Fitting the whole quotient uses the same math as unwrapping row by row. Desmos does all the rows at once.

The SAT gives exact values in these tables. So the regression isn't permission to accept a rough best fit. It's a way to solve for the function that matches every row exactly, so check that the function you get rebuilds every row of the table.

If $f$ is linear, write

f(x)=mx+b.

Then the quotient becomes

g(x)=\frac{mx+b}{x+c}.

With $x$ in x_1 and $g(x)$ in y_1, type:

y_1~(mx_1+b)/(x_1+c)

Put in the actual number for $c$ from the problem. The tilde tells Desmos to find the $m$ and $b$ that make the whole quotient match the table.

Here's the wrong way, and it's tempting. Don't type:

y_1~mx_1+b

That treats the table values as outputs of $f$ , but they're outputs of $g$ . Even when $f$ is a line, $g$ usually isn't. In the opening table, the values of $g$ fall from $6$ to $-3$ and then climb to $0$ , so no line passes through all three.

Check your understanding:

A table gives values of $h(x)=p(x)/(x-5)$ , where $p(x)=ax^2+bx+c$ . What custom regression keeps the whole relationship?

Which setup should you use?

Use row-by-row reconstruction when any one of these is true:

The denominator is simple, like $x+6$ , so each row takes one multiplication.
Seeing the hidden points would help. In the opening table, the row with $g(8)=0$ tells you right away that $f(8)=0$ too.
One row might already answer the question. If you only need $f(1)$ , that row gives $(1+6)(-3)=-21$ , and you're done.
The arithmetic by hand is shorter than setting up a fit.

Use a full quotient regression when all of these are true:

The question tells you what kind of function is hidden, like " $f$ is a linear function."
There are several unknowns to find, like $a$ , $b$ , and $c$ in a quadratic.
The quotient is easy to type without a slip.
You want $f$ ready to use. Once Desmos fits it, you can define $f$ on the next line and evaluate it.

Both setups should give you the same hidden function. If they don't, recheck the table labels, the signs, the denominator, and the function type.

Calculator loads as you approach

Type the quotient regression, define f, and check all three rows.

Stop when one row already gives the answer

Say

q(x)=\frac{r(x)+4}{2x-1},

and a table gives

q(2)=5.

The question asks for $r(2)$ . That one row is all you need. Plug it in:

5=\frac{r(2)+4}{2(2)-1}.

The denominator is $3$ , so multiply both sides by $3$ and subtract $4$ :

\begin{aligned} 15&=r(2)+4\\[1.4em] r(2)&=11. \end{aligned}

You never had to find the whole linear function $r$ , because the input the question asks about is already in the table.

Try it yourself:

Suppose the same table asked for $r(0)$ instead. Would the row at $x=2$ be enough? What else would you need?

It wouldn't be enough. The row at $x=2$ tells you only $r(2)$ . To find $r(0)$ , you need enough to pin down the whole line, like a second row at a different $x$ .

That's a habit worth keeping all over the SAT: stop as soon as you have what the question asks for, but don't guess at values the table can't tell you.

Keep the known terms around the hidden function

The hidden function isn't always the whole numerator.

Say

g(x)=\frac{f(x)+2x-5}{x-3}

and $f$ is linear. Write $f(x)=mx+b$ , and you get

g(x)=\frac{mx+b+2x-5}{x-3}.

The regression is:

y_1~(mx_1+b+2x_1-5)/(x_1-3)

Every known term stays right where the problem put it. Desmos fits only $m$ and $b$ , the two numbers that belong to $f$ .

To go row by row instead, solve for $f(x)$ . Multiply both sides by $x-3$ , then subtract $2x-5$ :

\begin{aligned} g(x)(x-3)&=f(x)+2x-5\\[1.4em] f(x)&=g(x)(x-3)-2x+5. \end{aligned}

The matching calculated column is:

y_1(x_1-3)-2x_1+5

Common mistake:

Fitting the whole numerator as $mx+b$ and dropping the known $2x-5$ . With the two rows in this calculator, that fit gives $4x-4$ , which is the whole numerator $f(x)+2x-5$ , not $f$ . The problem already gave you $2x-5$ , so keep it in and fit only the part that belongs to $f$ .

Check your understanding:

If $h(x)=(p(x)-4x+7)/(x+2)$ and $p(x)=ax+b$ , what custom regression do you type for a table of $h$ values?

Calculator loads as you approach

Try the regression or the calculated column. Both should give f(x) = 2x + 1.

Read the answer from the right function

Once the regression is done, define the hidden function on a new line. If the fit gives $m$ and $b$ , type:

f(x)=mx+b

Then turn the question into something you can type or read:

What the question asks about $f$	What to type or read
$y$ -coordinate of the $y$ -intercept	$f(0)$ , which is $b$ for $f(x)=mx+b$
$x$ -coordinate of the $x$ -intercept	Solve $f(x)=0$ , or click where the hidden graph crosses the $x$ -axis
Slope	Read $m$ from the fit
A value, like $f(7)$	Type `f(7)`
Vertex of a quadratic	Graph the quadratic you found, or use its coefficients
Minimum or maximum value	The vertex's $y$ -coordinate, once you've checked whether the parabola opens up or down

The quotient can have completely different intercepts and high or low points. In the opening problem, $f(0)=-24$ but $g(0)=-4$ . So name the function when you give your answer:

The $y$ -intercept of $f$ is $(0,-24)$ .

Saying "of $f$ " is one last check that you're answering on the right layer.

Calculator loads as you approach

f(0) = -24, but g(0) = -4. Change the inputs on the last two lines to compare other values of f and g.

The denominator restriction belongs to the quotient

g(x)=\frac{f(x)}{x-2},

then $g$ is undefined at

x=2,

because the denominator is zero there. So don't use a table row at that input, and don't claim the quotient relationship holds there.

The hidden function $f$ , though, may still be defined at $x=2$ . Take

f(x)=x^2+x-6=(x-2)(x+3).

Then

f(2)=0,

even though

\frac{f(2)}{2-2}

is undefined. The gap belongs to the quotient $g$ . It doesn't automatically carry over to the polynomial $f$ .

For holes, asymptotes, and canceled factors, look back at Restrictions, piecewise functions, and rational expressions.

Check your understanding:

If $h(x)=p(x)/(x+4)$ and $p$ is a polynomial, which is guaranteed: $h(-4)$ is undefined, $p(-4)$ is undefined, or both?

Guided example: Recover a hidden line from a bigger quotient

Worked example

The table gives two values of $x$ and their corresponding values of $g(x)$ , where

g(x)=\frac{f(x)+2x-5}{x-3}

and $f$ is a linear function.

$x$	$g(x)$
$1$	$4$
$5$	$-1$

Which choice is the $y$ -coordinate of the $y$ -intercept of the graph of $y=f(x)$ in the $xy$ -plane?

A
$-\frac{7}{2}$
B
$-\frac{11}{2}$
C
$-\frac{9}{4}$
D
$-\frac{9}{2}$

Step 1

Name the layers

The table gives values of $g$ , but the question asks for the $y$ -intercept of $f$ . The denominator also rules out $x=3$ for the quotient. Neither row uses that input, so both rows are safe to use.

Step 2

Write f as a line

Since $f$ is linear, write

f(x)=mx+b.

Put that into the whole quotient:

g(x)=\frac{mx+b+2x-5}{x-3}.

The known $2x-5$ stays in the numerator.

Step 3

Fit the whole quotient

The two rows are already in the table. Two rows are enough here, because there are only two unknowns, $m$ and $b$ . Type:

y_1~(mx_1+b+2x_1-5)/(x_1-3)

Desmos shows the decimals

m=-0.5

and

b=-4.5.

As fractions, those are $-\frac12$ and $-\frac92$ . Check those exact values in the original quotient before you trust them.

Step 4

Answer from f, then check both rows

The hidden function is

f(x)=-\frac12x-\frac92.

f(0)=-\frac92,

and the $y$ -coordinate of the $y$ -intercept is

\boxed{-\frac92}.

The answer is D.

Now check it in the original quotient:

\frac{f(1)+2(1)-5}{1-3} = \frac{-5+2-5}{-2} =4

and

\frac{f(5)+2(5)-5}{5-3} = \frac{-7+10-5}{2} =-1.

Both match the table.

Common mistake:

Clicking where the fitted curve crosses the $y$ -axis. That curve is $g$ , and it crosses at $\frac{19}{6}$ . The question asks about $f$ , so evaluate $f(0)$ once you’ve found $f$ .

Calculator loads as you approach

At step 3, type the quotient regression, define f, and find f(0).

Check the original relationship

A clean regression result doesn't finish the problem. You still need to know that the function you found rebuilds the values in the table.

For each row:

Evaluate the hidden function you found.
Put that value back into the original numerator.
Divide by the original denominator.
Check that you get the value in the table.
Make sure the denominator isn't zero.

For the opening problem, $f(x)=3x-24$ . At $x=1$ :

\frac{f(1)}{1+6} = \frac{-21}{7} =-3,

which matches the table.

You don't need to write out every check on the SAT. Check one or two rows when the setup is easy to mistype.

Calculator loads as you approach

Change the input in one check, compare the result with the matching value of g, then reset.

Finish the solution

Recover the hidden y-intercept

Finish the solution

The table gives four values of $x$ and their corresponding values of $g(x)$ , where

g(x)=\frac{f(x)}{x+2}

and $f$ is a linear function.

$x$	$g(x)$
$-6$	$1$
$-4$	$-2$
$1$	$8$
$4$	$6$

What is the $y$ -coordinate of the $y$ -intercept of the graph of $y=f(x)$ ?

First steps

The table is already in the calculator. Fit the whole quotient, define $f$ , and submit the $y$ -coordinate of its $y$ -intercept.

Let $f(x)=mx+b$ .
Keep the whole relationship: $g(x)=(mx+b)/(x+2)$ .
Fit that relationship to the table.
Define $f$ and evaluate $f(0)$ .

Finish it

Your answer

Calculator loads as you approach

Fit the whole quotient, then define f and find f(0).

Practice problems

Your turn. Try each problem before you open its hint, and fit the whole quotient only when that's quicker than working by hand.

Stop at the row you need

Practice problem

The table gives three values of $x$ and their corresponding values of $q(x)$ , where

q(x)=\frac{r(x)+4}{2x-1}

and $r$ is a linear function.

$x$	$q(x)$
$-1$	$-2$
$2$	$5$
$5$	$\frac83$

What is the value of $r(2)$ ?

Your answer

Calculator loads as you approach

Look for the row you need before you set up a regression.

Recover a hidden x-intercept

Practice problem

The table gives four values of $x$ and their corresponding values of $g(x)$ , where

g(x)=\frac{f(x)}{x+4}

and $f$ is a linear function.

$x$	$g(x)$
$-8$	$-8$
$-6$	$-13$
$0$	$2$
$1$	$1$

What is the $x$ -coordinate of the $x$ -intercept of the graph of $y=f(x)$ ?

Your answer

Calculator loads as you approach

Fit the whole quotient or go row by row, then find where f crosses the x-axis.

Keep the known numerator terms

Practice problem

The table gives three values of $x$ and their corresponding values of $g(x)$ , where

g(x)=\frac{f(x)-3x+2}{x+1}

and $f$ is a linear function.

$x$	$g(x)$
$-5$	$-8$
$1$	$1$
$3$	$-2$

What is the value of $f(-2)$ ?

Your answer

Calculator loads as you approach

Keep -3x + 2 in the numerator while you fit f.

Recover a hidden quadratic

Practice problem

The function $q$ is defined by

q(x)=\frac{p(x)-5}{x+1},

where $p$ is a quadratic polynomial. The table shows three values of $x$ and the corresponding values of $q(x)$ .

$x$	$q(x)$
$2$	$3$
$4$	$7$
$-4$	$-1$

What are the coordinates of the vertex of the graph of $y=p(x)$ in the $xy$ -plane?

Calculator loads as you approach

Put a quadratic in the numerator, then find the vertex of p.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Name the layers. Say which function the table gives and which one the question asks about.
Note the restriction. Where the denominator is zero, the quotient relationship doesn't hold.
Unwrap. Solve the relationship for the hidden value and work it out row by row.
Model. Put the hidden function's type into the whole quotient before you run a regression.
Keep known terms. Every term around the hidden function stays where the problem put it.
Stop early. If one row already gives the hidden value you need, don't fit the whole function.
Read from the right function. Take the intercept, value, slope, or vertex from the hidden function, not automatically from the table's function.
Check. Put the function you found back into the original quotient and make sure it rebuilds the table.

Next lesson

Exactness, rounding, and output verification

Decide when a calculator decimal is final, when you can recover an exact form, and how to match the precision the question asks for.

Start next lesson

Practice

Practice this lesson

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

Start practice