Custom regression with multiple conditions

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Learn to count parameters and independent conditions, fit exact custom models with matched lists or tables, and verify the result the SAT asks for.

Difficulty: Advanced
Estimated time: 45 minutes
Domains: Algebra Advanced Math

What you’ll learn

Count the unknown parameters and the independent conditions before you fit anything.
Tell a condition that adds new information from one that repeats another.
Put mixed conditions on one line as two lists of the same length, paired by position.
Use a table when every condition is a point.
Type the model exactly as given, with only the intended parameters left free.
Treat Desmos's decimals as likely values and confirm them exactly before you trust them.
Answer what the question asks for and check every original condition.
Spot when substitution or graphing an ordinary system is faster.

Prerequisites

Custom regression for one unknown parameter0%
Lesson 18: keep a given model as written, fit one unknown parameter from one condition, and verify the result.

Why this matters on the SAT

Fit several exact conditions and keep the model whole

Some SAT questions give you a function with a few unknown constants, called parameters, and enough exact facts to find them all. You've already used custom regression for one unknown parameter, where one condition found one parameter. Now there are several of each, and one regression line can still handle them together.

Here's a typical question.

SAT example

The quadratic function $q$ is defined by

q(x)=a(x-1)^2+b,

where $a$ and $b$ are constants. If $q(0)=12$ and $q(4)=0$ , what is the maximum value of $q$ ?

Your answer

Fast Desmos solution

Each condition says a known output equals the model at some input. Put the two outputs in one list and the two model expressions in another, on a single regression line:

[12,0]~[a(0-1)^2+b,a(4-1)^2+b]

Desmos reports $a=-1.5=-\frac32$ and $b=13.5=\frac{27}{2}$ . The model is in vertex form, and $a$ is negative, so the parabola opens down and its highest point is the vertex, at $x=1$ . The maximum is

q(1)=b=\boxed{13.5}.

The lists don't change the model. They tell Desmos that the same $a$ and $b$ have to make both conditions true.

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Two function values pin down a and b, and q(1) gives the maximum.

When should you use this method?

Reach for this when the question gives you a specific model with several unknown parameters, plus at least as many independent exact conditions as there are parameters. One set of parameter values has to make every condition true at once.

Then pick the setup that fits the conditions.

Use custom regression when…

Table custom regression fits when every condition is a point or a function value, like $f(1)=7$ and $f(3)=31$ .
Matched-list custom regression fits when the conditions come in different forms, like a function value $f(2)=9$ together with a fact about the parameters, such as $a=2b$ .

Try a shorter method when…

Substitution wins when one condition hands you a parameter right away or leaves an easy equation. For $f(x)=a(x-2)^2+b$ , the fact $f(2)=3$ gives $b=3$ in one step.
Ordinary system graphing wins when the problem is just a system in $x$ and $y$ , like $y=2x+1$ and $y=-x+7$ . Type both equations with equals signs and read where they cross.

This isn’t the same job as Desmos regression from a supplied table, where Desmos uses an automatic model to summarize data. Here the question gives you the exact model and exact conditions, so you type that model as written. It’s also different from parameterized points and answer choices, where a parameter moves a point through infinitely many positions instead of being fitted from conditions.

In the models here, each parameter stays simple: it multiplies a known expression or is added to another term. It never sits in an exponent, a denominator or a root, and it’s never multiplied by another fitted parameter. Identities, where two expressions have to match at every input, come in the next lesson, find constants in equivalent expressions. Circles get their own regression in from geometry to coordinates to circle regression, and noisy data, along with cases where Desmos’s output can’t be treated as an exact solution, in the lesson on regression failure modes.

Check your understanding:

You need to solve $2x+y=11$ and $x-y=1$ . Would you start with matched-list regression or ordinary system graphing?

Count parameters and independent conditions

Before you type anything, count two things: the letters Desmos has to find, and the conditions that tell you something new. You need at least this many:

\text{number of independent conditions} \ge \text{number of unknown parameters}.

Matching the numbers isn't always enough, though, because two conditions can say the same thing. Take

f(x)=a(x-2)^2+b

with

f(0)=7 \quad\text{and}\quad f(4)=7.

The model has two unknown parameters, $a$ and $b$ . Plug in $x=0$ : $a(0-2)^2+b=7$ , so the first condition gives

4a+b=7.

Plug in $x=4$ : $a(4-2)^2+b=7$ , so the second condition gives

4a+b=7.

It's the same equation twice. These conditions are redundant: the two points sit at the same height, one on each side of the vertex at $x=2$ , so they carry only one independent equation. Lots of pairs fit it, like $a=1,\ b=3$ and $a=2,\ b=-1$ . In fact infinitely many pairs $(a,b)$ work, so the function isn't pinned down.

Now add a condition such as $f(2)=3$ . It gives

b=3,

which is new information, independent of $4a+b=7$ . With two independent equations, both parameters are fixed: $b=3$ , and then $4a+3=7$ gives $a=1$ .

So when you count, count new facts, not stated facts.

Check your understanding:

For $h(x)=mx+c$ , you’re told $h(0)=4$ , that the $y$ -intercept is $4$ , and that $h(3)=10$ . How many independent conditions are there, and are they enough to find $m$ and $c$ ?

Put mixed conditions on one matched-list line

When the conditions aren't all input-output points, write each one as "known value equals model expression." Then put the known values in one list and the expressions in a second list, in the same order:

[known value 1,known value 2]~[model expression 1,model expression 2]

Desmos pairs the entries by position. The first entry on the left goes with the first entry on the right, the second with the second, and every pair has to work with the same fitted values of the parameters. So the two lists need the same length, and the order matters: keep each known value lined up with its own expression.

In the opening example, the line

[12,0]~[a(0-1)^2+b,a(4-1)^2+b]

means

\begin{aligned} 12&=a(0-1)^2+b,\\[1.4em] 0&=a(4-1)^2+b. \end{aligned}

Desmos fits both equations together, which is why it finds one shared pair, $a=-\frac32$ and $b=\frac{27}{2}$ . One line, one shared set of values.

Common mistake:

Each regression line is its own fit, so separate lines don’t reliably share one set of parameter values. Put every condition you need on one matched-list line.

Check your understanding:

Write $2a+b=9$ and $a-3b=-5$ as one matched-list regression line. What does the order of the lists mean?

Once you've written your line, check it in the calculator. You should see one value for $a$ and one for $b$ . Then change either number on the left and watch both fitted values move together. Reset the calculator when you're done.

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One matched-list line fits one shared pair of values for a and b.

Use a table when every condition is a point

Worked example

Suppose

f(x)=a\cdot2^x+b,

where $a$ and $b$ are constants. If $f(1)=7$ and $f(3)=31$ , what is the value of $b$ ?

Both conditions are function values, so they're the points $(1,7)$ and $(3,31)$ . Put those points in a table, then type the exact model:

y_1~a*2^(x_1)+b

Behind the scenes, the two rows say

\begin{aligned} 2a+b&=7,\\[1.4em] 8a+b&=31. \end{aligned}

Subtract the first equation from the second and you get $6a=24$ , so $a=4$ . Then $2(4)+b=7$ , so $b=-1$ . The question asks for $b$ , so the answer is

\boxed{-1}.

This isn't Desmos's automatic exponential regression. In the given model, the base is fixed at $2$ , and $b$ shifts the graph up or down. An automatic exponential template can bring in a different set of free parameters. Custom regression keeps exactly

a\cdot2^x+b.

Try it yourself:

Change the second row to input $2$ and output $15$ . Before you look, predict: will $a=4$ and $b=-1$ still fit both points? Then reset the example.

Check your understanding:

Table or matched lists? (1) A stated model with $p(2)=5$ and $p(6)=17$ . (2) A stated model with $p(2)=5$ , where the parameters also satisfy $a+b=4$ .

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The two table rows pin down a = 4 and b = -1, with the base kept at 2.

Keep only the intended parameters free

Before you fit, look at every letter that doesn't have a value yet.

For

f(x)=a\cdot2^x+b,

the parameters you want Desmos to find are $a$ and $b$ . The base $2$ is fixed by the problem. In the table regression:

x_1 and y_1 hold the paired table values.
$a$ and $b$ stay undefined, so Desmos can fit them.
Every other number and operation stays exactly as given.

Don't swap the fixed base for another letter. A model such as

y_1~a*c^(x_1)+b

has three unknown parameters, $a$ , $b$ and $c$ , but only two point conditions. It's no longer the problem you were given.

Also check that you haven't given $a$ or $b$ a value on an earlier line. If you have, Desmos treats it as fixed and won't fit it. Delete the old definition, or switch to a new letter and use it everywhere.

Check decimals against the exact conditions

Desmos gives you decimals. A value like $-1.5$ suggests $-\frac32$ , but you can only call the fraction exact once it makes the original equations true exactly.

For the opening example:

\begin{aligned} q(0) &=-\frac32(0-1)^2+\frac{27}{2} =12,\\[1.4em] q(4) &=-\frac32(4-1)^2+\frac{27}{2} =0. \end{aligned}

Both original conditions hold, so the exact values are confirmed.

An extra exact condition makes a good check, but it can't stand in for a missing independent one. Three copies of $4a+b=7$ are still one equation.

If even one required exact condition fails, throw out the fitted values and fix the setup before you answer. Values that satisfy only some of the conditions don't solve the problem you were given.

Example: Match mixed conditions

Worked example

The function $f$ is defined by

f(x)=ax+b,

where $a$ and $b$ are constants. The function $g$ is defined by

g(x)=f(2x).

If $g(3)=5$ and $a+b=0$ , what is the value of $f(0)$ ?

A
$-1$
B
$0$
C
$1$
D
$2$

Step 1

Translate the condition on g

Careful here: the condition $g(3)=5$ is about $g$ , not $f$ . So first write $g$ using the rule for $f$ , with $2x$ in place of $x$ :

g(x)=f(2x)=a(2x)+b.

That makes

g(3)=a(2\cdot3)+b=6a+b=5.

The second condition, $a+b=0$ , is already about the parameters, so it needs no translating.

Step 2

Pair the conditions by position

One condition is a function value and the other is a fact about the parameters, so use matched lists, not a table:

[5,0]~[a(2*3)+b,a+b]

Desmos reports $a=1$ and $b=-1$ .

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The first pair is g(3) = 5, and the second is a + b = 0.

Step 3

Answer what the question asks

Desmos found $a$ and $b$ , but the question wants $f(0)$ . Go back to $f$ :

f(0)=a(0)+b=b=-1.

The answer is A.

Step 4

Check both original conditions

With $a=1$ and $b=-1$ ,

g(x)=1(2x)-1=2x-1.

Now check both conditions:

\begin{aligned} g(3)&=2(3)-1=5,\\[1.4em] a+b&=1+(-1)=0. \end{aligned}

Both hold, and $f(0)=-1$ answers the question.

Common mistake:

It’s tempting to put both conditions in a table. But $a+b=0$ has no input and output, so there’s no point to enter. Keep it as the second entry in the matched lists.

Read the output, answer, and check

Every multiple-condition regression ends with the same three jobs:

Read the fitted parameters. Treat the decimals Desmos shows as likely exact values, not confirmed ones yet.
Answer the question. The SAT may ask for a function value, an intercept, a maximum or an expression rather than a parameter.
Check every condition. Plug the fitted values into each original condition, not only one of them. If any exact condition fails, fix the setup.

In the guided example, $a=1$ and $b=-1$ were the fitted values, $f(0)=-1$ was the answer, and $g(3)=5$ and $a+b=0$ were the checks.

Finish the solution

The matched-list regression below has already found $a$ and $b$ . Use them to find the function value the question asks for.

Finish the solution

The quadratic function $q$ is defined by

q(x)=a(x-3)^2+b,

where $a$ and $b$ are constants. If $q(0)=23$ and $2a-b=-1$ , what is the value of $q(5)$ ?

First steps

Turn $q(0)=23$ into $a(0-3)^2+b=23$ .
Keep the parameter fact $2a-b=-1$ as it is.
Fit both conditions together with [23,-1]~[a(0-3)^2+b,2a-b].

Finish it

Your answer

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Desmos has already fitted a and b. Add q(5) to finish.

Practice problems

Start each one by counting the unknown parameters and the independent conditions. Then pick the shortest setup: a table, matched lists, substitution or ordinary system graphing. Not every problem here needs a regression.

Practice problem

The linear function $f$ is defined by

f(x)=ax+b,

where $a$ and $b$ are constants. If $f(5)=15$ , the slope of the graph of $f$ is $3$ less than its $y$ -intercept, and $f(0)=5$ , what is the value of $f(-2)$ ?

Your answer

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Try one matched-list line that holds all three conditions.

Practice problem

The function $C$ models the cost, in dollars, of processing $g$ grams of a material. The model is

C(g)=a+b(g-5),

where $a$ and $b$ are constants. The graph of $C$ passes through $(3,52)$ and $(9,88)$ . What is the value of $a$ ?

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Put the two cost points in a table and fit the model exactly as given.

Practice problem

The quadratic function $f$ is defined by

f(x)=a(x-2)^2+b,

where $a$ and $b$ are constants. If $f(1)=7$ and $f(5)=31$ , what is the value of $f(8)$ ?

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Two points in a table, vertex form kept as given, then f(8).

Practice problem

The quadratic function $f$ is defined by

f(x)=a(x-4)^2+b,

where $a$ and $b$ are constants. If $f(4)=7$ and $f(6)=19$ , which approach is shortest for determining $f(8)$ ?

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Pick your method before you type, then use Desmos to check f(8).

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Recognize: The question gives a specific model with several unknown parameters and enough independent exact conditions. Count new facts, not stated facts.
Choose: Use a table when every condition is a point, and one matched-list line when the conditions come in different forms. Go with substitution or ordinary system graphing when either is shorter.
Fit: Type the model exactly as given, keep the lists the same length and in matching order, and leave only the intended parameters free.
Verify: Treat the decimals as likely values, answer what the question asks, and check every original condition exactly.

Next lesson

Find constants in equivalent expressions

Fit both sides of an identity over a list of inputs to find its unknown constants.

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