Meet every condition

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Turn a dense SAT function question into an ordered Desmos model, choose legal witness values, and verify that the result it asks for stays fixed.

Difficulty: GOAT Mode
Estimated time: 50 minutes
Practice: Math question bank

What you’ll learn

Sort a crowded question into a condition stack: fixed values, free constants, values that depend on them, and the target.
Pick a legal witness, one set of values that makes every condition true.
Type definitions and outputs in Desmos in dependency order, so each line comes after what it uses.
Use a second witness to catch mistakes, then show with algebra that the target is invariant.
Tell when to stop using witness values: when the target changes, or when you need to test every possible value.

Prerequisites

Evaluate, combine, and compose functions0%
Lesson 5: define functions and evaluate named outputs in Desmos.

Why this matters on the SAT

Sort out the conditions before you calculate

Some of the hardest SAT function questions pile on constants, inequalities, named outputs, and one messy expression to find. We'll call that expression the target. Often you can't pin down the constants, and yet the target can still have exactly one answer.

Here's the move: choose values that make every condition true, and build the whole thing in Desmos, one line at a time, in order.

SAT example

The function $q$ is defined by

q(x)=ax+b,

where $a>b$ and $b>87$ . Let

H=q(639) \qquad\text{and}\qquad K=q(621).

What is the value of

H-K-18a?

Your answer

Fast Desmos solution

Nothing in the question pins down $a$ and $b$ . So pick values that follow the rules: $b$ has to be more than $87$ , and $a$ has to be more than $b$ . On separate lines, type b=88 and a=89.

Then type q(x)=a*x+b, H=q(639), K=q(621), and H-K-18a, each on its own line.

Desmos returns $0$ .

Now switch to $b=90$ and $a=92$ , which follow the rules too. The target is still $0$ .

Try it yourself:

Change the calculator to $(a,b)=(92,90)$ . Before you read the target, check that $92>90$ and $90>87$ . Then reset the example.

Two matching answers are a good sign, but they don't prove the answer is always $0$ . The algebra does:

\begin{aligned} H-K-18a &=(639a+b)-(621a+b)-18a\\[1.4em] &=18a-18a\\[1.4em] &=0. \end{aligned}

The $b$ 's cancel, and the $18a$ you subtract wipes out what's left. So every legal pair gives $0$ , not only the two you tried. That's the idea to hold on to: Desmos checks, algebra proves.

Calculator loads as you approach

Each line comes after the values it uses. Switch to another legal pair, and the target stays 0.

When should you pick witness values?

Pick witness values when all of these are true.

The question defines one or more functions, like $q(x)=ax+b$ .
Some constants are limited only by inequalities, like $a>b$ and $b>87$ , or by simple relationships.
There isn't enough information to work out every constant exactly.
It asks for a mix of outputs and constants, like $H-K-18a$ , that might come out the same whichever legal values you choose.

Some questions look similar but need a different approach. Don't pick witness values in these cases.

The question asks for a constant itself, like "What is the value of $a$ ?"
Another exact condition settles the constant. For example, $q(0)=90$ tells you $b=90$ .
The target changes when you change a legal value, so there's no single answer to report.
The question asks for every solution, the least or greatest possible value, or how many integers work.
The conditions pin down several constants exactly. Then solve the conditions as a system, or try custom regression with multiple conditions, and check that every condition holds.

Check your understanding:

A question says $a>b>10$ , defines $f(x)=ax+b$ , and asks for $f(8)-f(3)-5a$ . Nothing else pins down $a$ or $b$ . Would you pick witness values here? If so, what answer should every legal witness give?

Build a condition stack

Give every fact one job

A condition stack is a list, in order, of everything that has to be true before you work out the target.

Here's the stack for the SAT example:

Layer	From the question	What it means
Fixed values	$639$ , $621$ , and $18$	Copy these exactly from the question.
Free constants	$a>b$ and $b>87$	$a$ and $b$ can vary, but only within these rules.
Function definition	$q(x)=ax+b$	Every output below uses this rule.
Dependent values	$H=q(639)$ and $K=q(621)$	$H$ and $K$ come from the function and the constants.
Target	$H-K-18a$	Work this out last, once everything above it is defined.

The order matters, because each layer is built from the ones above it.

Before you type anything, ask yourself four questions:

Which numbers does the question give me outright?
Which constants can vary, and what rules limit them?
Which values come from a function or an earlier definition?
What exact expression do I need to find?

Check your understanding:

In the SAT example, is $H$ a free constant or a dependent value? What has to be defined before $H$ ?

Common mistake:

A small condition, like a sign, an order, an integer rule, or a limit on the inputs, can decide which value works. Write down every condition before you calculate, even one that never shows up in the final expression.

Choose one legal witness

Free doesn't mean anything goes

A witness is one set of values that makes every condition in the question true. We call it legal to stress that point: it has to pass every single rule, not most of them.

In the SAT example, the rules are $a>b$ and $b>87$ . The pair $(a,b)=(89,88)$ is legal, because $89>88$ and $88>87$ .

These pairs aren't legal:

$(88,89)$ breaks $a>b$ , because $88$ is less than $89$ .
$(88,87)$ breaks $b>87$ , because $87$ isn't more than $87$ .
$(88,88)$ breaks $a>b$ . The sign is strict, so equal values don't count.

Pick simple values right past the boundary, like $88$ and $89$ , so the arithmetic stays easy. But avoid $0$ and $1$ when they could hide a term, a sign, an exponent, or a coefficient. Multiplying by $0$ wipes a term out, and multiplying by $1$ leaves it unchanged, so a typo in that spot would slip past you.

Some rules keep the math itself working. If the target contains $\frac{1}{a-b}$ and the question says $a>b$ , then choosing $a=b$ means dividing by $0$ , and the expression is undefined. Sometimes an answer does hold outside the legal values too, but you can't count on that, so check every condition before you use a witness.

Check your understanding:

The rule is $c>d>20$ . Which pair is legal: $(c,d)=(21,22)$ , $(22,21)$ , or $(21,21)$ ?

Build in dependency order

Put the new name on the left

When you define an output in Desmos, put the new name on the left. Type H=q(639) and K=q(621) on separate lines.

Now $H$ and $K$ are values you can reuse in later lines. If you type q(639)=H instead, Desmos reads the line as an equation, not a direct definition, and it's harder to see what depends on what.

Here's a clear order, one line each: b=88, a=89, q(x)=a*x+b, H=q(639), K=q(621), and last, H-K-18a.

To be fair, Desmos doesn't need this order. It connects the definitions and updates everything that depends on them, wherever the lines sit in the list. The order is for you: it makes the model easy to check and cuts down on copying mistakes. If a name stays undefined, make sure its definition is there and has no error.

Try it yourself:

Delete the last three lines, then type H, K, and the target back in, each after what it uses. Reset the calculator when you’re done.

Check your understanding:

Put these lines in dependency order, so each one comes after what it uses: T=U+V-32c, U=p(d+4), p(x)=c(x-d)^2+7, c=2, d=51, and V=p(d-4).

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Each name comes after the constants or function it uses.

Example: Prove a symmetric quadratic stays fixed

One witness that works isn't proof

An expression is invariant when it gives the same value for every legal witness the question allows.

Here's the three-step check:

Work out the target with one legal witness.
Switch to a clearly different legal witness and work it out again.
Explain why the result didn't change, using cancellation, symmetry, or another exact relationship.

If the target does change, stop and find out why. One of these is going on:

The question doesn't pin down one answer.
A condition is missing from your model.
The new witness isn't legal.
Something was typed in wrong.

Worked example

The function $p$ is defined by

p(x)=c(x-d)^2+7,

where $c>0$ and $d>50$ . Let

U=p(d+4) \qquad\text{and}\qquad V=p(d-4).

What is the value of

U+V-32c?

A
$0$
B
$7$
C
$14$
D
$32$

Step 1

Build the stack

The constants $c$ and $d$ can vary, as long as $c>0$ and $d>50$ . The function uses $c$ and $d$ , and $U$ and $V$ use the function.

So start with a legal witness, like

c=2 \qquad\text{and}\qquad d=51.

Why $c=2$ and not $c=1$ ? With $c=1$ , forgetting to multiply by $c$ somewhere wouldn't change the result, so that mistake would stay hidden.

Step 2

Define the outputs and target

Type c=2, d=51, p(x)=c*(x-d)^2+7, U=p(d+4), V=p(d-4), and U+V-32c on separate lines, in that order.

Desmos returns $14$ .

Step 3

Change the witness, then explain

In the same calculator, change both constants:

c=3 \qquad\text{and}\qquad d=60.

The target is still $14$ .

Now for the why. Both inputs sit $4$ away from $d$ , one on each side, so the squared part is $16$ either way:

\begin{aligned} U&=c(4)^2+7=16c+7,\\[1.4em] V&=c(-4)^2+7=16c+7. \end{aligned}

U+V-32c =(16c+7)+(16c+7)-32c =14.

The answer is C.

Common mistake:

Two matches can still fool you. Suppose the target were $U+V-30c$ , and your second witness changed only $d$ . You’d get $18$ both times, yet the target is really $2c+14$ , which changes as soon as $c$ does. So make the second witness clearly different, and let the algebra have the final word.

Calculator loads as you approach

Test c = 2, d = 51, then change to c = 3, d = 60 and compare the target.

Stop when witness values are unsafe

A changing target is a warning sign

Here's what it looks like when witness values let you down. Suppose

q(x)=ax+b, \qquad a>b>2,

and

H=q(5), \qquad K=q(1).

Then $H=5a+b$ and $K=a+b$ , so

H-K=4a.

That value moves whenever $a$ moves. If you picked one legal witness and reported its output, you'd be making up a single answer that the question never gave you.

Try it yourself:

Change a=4 to a=6. Then change only $b$ , keeping it legal. Which constant controls $H-K$ ? Reset the calculator when you’re done.

Check your understanding:

Using $(a,b)=(4,3)$ gives $H-K=16$ . Using $(a,b)=(6,3)$ gives $H-K=24$ . What should you conclude?

Some questions need you to test every possible value

The same caution goes for questions that ask for a least or greatest value, every solution, or the exact value of a constant. There you solve exactly, or use candidate filtering: test every possible value and check each one against every condition.

Here's why one witness can't do that job. Say $k$ must be the least positive integer that fits several relationships. Finding one $k$ that fits shows that it works. It doesn't show that no smaller value works.

Common mistake:

One value that works shows it’s possible, not that it’s the smallest. Before you call a value the least, check every smaller value the question allows.

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Change a from 4 to 6, and the target jumps from 16 to 24. It isn’t invariant.

Finish the solution

Your turn to finish one. The first legal witness and the function are already in the calculator.

Finish the solution

The function $f$ is defined by

f(x)=m(x-n)+4,

where $m>n>10$ . Let

A=f(n+6) \qquad\text{and}\qquad B=f(n-2).

What is the value of

A-B-8m?

First steps

The pair $(m,n)=(12,11)$ is legal, since $12>11>10$ .
The function f(x)=m*(x-n)+4 is already typed in.
Define $A$ and $B$ using the function.
Work out the target once both outputs are defined.

Finish it

Your answer

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Add A, B, and the target, then switch to another legal pair.

Practice problems

Four to try on your own. Build the stack first, and give each problem a real try before you open its hint.

Practice problem

The function $g$ is defined by

g(x)=r(x-s)^2+5,

where $r>s>0$ . Let

U=g(s+5) \qquad\text{and}\qquad V=g(s-1).

What is the value of

U-V-24r?

Your answer

Calculator loads as you approach

Pick a legal pair, define g, U, and V, then try a second pair.

Practice problem

The function $h$ is defined by

h(x)=ax^2+bx+6,

where $a>b>10$ . Let

P=h(4) \qquad\text{and}\qquad Q=h(-4).

What is the value of

P+Q-32a?

Calculator loads as you approach

Pick a legal pair, compare h(4) with h(-4), then work out the target.

Practice problem

The function $P$ is defined by

P(t)=ca^t+5,

where $c>a>1$ . Let

H=P(5) \qquad\text{and}\qquad K=P(3).

What is the value of

H-a^2K+5a^2?

Your answer

Calculator loads as you approach

Pick c greater than a, and a greater than 1, then define H, K, and the target.

Practice problem

The function $f$ is defined by

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

where $a>0$ , $b>10$ , and $c>0$ . Which expression is equivalent to

f(b+3)+f(b-3)?

Calculator loads as you approach

Try two legal triples and notice that the target changes. Then simplify it.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Write down every condition, then sort it: fixed values, free constants, dependent values, and the target.
Pick a legal witness, and type definitions in dependency order so the model is easy to check.
Try a clearly different second witness to catch mistakes. It checks the answer, but it doesn't prove it.
Prove the target is invariant with cancellation, symmetry, or another exact relationship. Desmos checks, algebra proves.
If the target changes, or the question asks for a least, greatest, integer, or complete set of answers, test every possible value or solve exactly.

Next lesson

Hidden functions inside quotients

Recover useful function information when the question hides it inside a rational expression.

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