Restrictions, piecewise functions, and rational expressions

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Difficulty: Intermediate
Estimated time: 55 minutes
Domains: Advanced Math
Techniques: RestrictionsDomainPiecewise-functionsRational-expressionsSolution-checking

What you’ll learn

Add domain restrictions to a graph with curly braces.
Enter and evaluate a piecewise function in Desmos.
Record denominator exclusions before simplifying a rational expression.
Distinguish an x-intercept, a hole, and a vertical asymptote.
Check possible solutions in the original rational or radical equation.
Choose when a graph helps and when a short algebraic check is faster.

Prerequisites

Why this matters on the SAT

The equation is not the whole story

An input can satisfy your simplified algebra and still fail the original problem. A denominator can equal zero. A piecewise rule can apply only on one interval. Squaring both sides can produce a value that was never a solution.

On the SAT, these conditions often decide between the tempting answer and the correct one.

SAT example

Which choice is the $x$ -coordinate of the $x$ -intercept of the graph of

y=\frac{x^2-1}{x^2-4x+3}

in the $xy$ -plane?

A
$-1$
B
$1$
C
$3$
D
$0$

Fast Desmos solution

Graph the function and click its marked $x$ -intercept. Desmos shows $(-1,0)$ , so the answer is A.

Factoring the denominator gives

x^2-4x+3=(x-1)(x-3),

so $x=1$ and $x=3$ are excluded. The graph has a hole at $x=1$ after a common factor cancels and a vertical asymptote at $x=3$ . Neither excluded input can be an $x$ -intercept.

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Click the marked x-intercept. Then inspect the graph near x=1 and x=3.

Choose the right approach

A function or equation comes with conditions. Keep those conditions attached from the original problem to the final answer.

The domain is the set of inputs for which an expression or function is defined.

Desmos can display many of those conditions directly, but you still need to read them before entering the math.

Choose the method that keeps every condition visible with the least work.

Use a Desmos restriction when…

the prompt gives an interval such as $0\le x\le5$ ;
only part of a graph makes sense in context;
a function uses different rules on different intervals;
you need an endpoint, extreme, or intersection on an allowed domain.
graphing the original sides will filter invalid candidates.

Write the restrictions first when…

a variable appears in a denominator;
you plan to cancel a common factor;
you square both sides or clear denominators;
a radical or logarithm limits the domain; or
an exact exclusion is easier to identify algebraically than visually.

Use the shorter method. If one substitution rejects a candidate immediately, do it. If a restricted graph exposes the valid interval or intersection at once, use the graph.

Check your understanding:

A problem asks for the maximum of a quadratic on $0\le x\le5$ . Another asks you to simplify a rational expression with denominator $x-4$ . What should you do first in each?

1. Restrict a graph with curly braces

Suppose

h(x)=-(x-3)^2+9

is defined only for

0\le x\le5.

Put the condition at the end of the complete equation:

y=-(x-3)^2+9{0<=x<=5}

Desmos graphs only the allowed part of the parabola. The maximum is $9$ at $x=3$ . Both endpoints are included because the inequalities use $\le$ .

Use < or > when an endpoint is excluded.
Use <= or >= when an endpoint is included.
Use a compound inequality such as 0<=x<=5 to restrict both ends at once.

An included endpoint remains part of the graph. An excluded endpoint is omitted.

Try it yourself:

Edit the right side of the restriction from x<=5 to x<5. The endpoint at $x=5$ disappears because $5$ is no longer allowed.

Check your understanding:

For the graph $y=2x+1\{-2<x\le3\}$ , which endpoint is included?

Common mistake:

Using parentheses instead of a restriction. Parentheses group algebra. Curly braces tell Desmos which points to keep. Put the restriction after the complete equation.

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The curly-brace condition removes every point whose x-coordinate is outside 0≤x≤5.

2. Enter a piecewise function

Consider

p(x)= \begin{cases} 2x+3, & x<0\\ x^2+1, & x\ge0. \end{cases}

You can graph the pieces on separate lines:

y=2x+3{x<0}
y=x^2+1{x>=0}

Separate lines are convenient when you want to edit each branch. A compact function definition is better when you also want to evaluate $p(x)$ :

p(x)={x<0:2x+3,x>=0:x^2+1}

The condition comes before the colon, and the rule used under that condition comes after it.

Read the condition before the formula. For this function,

p(-2)=2(-2)+3=-1

because $-2<0$ ,

p(0)=0^2+1=1

because $0\ge0$ , and

p(3)=3^2+1=10.

Check your understanding:

Define $r(x)=3-x$ for $x<1$ and $r(x)=x^2$ for $x\ge1$ in one Desmos line. Then find $r(-1)+r(2)$ .

Common mistake:

Ignoring the boundary symbol. At $x=0$ , the first rule does not apply because it uses $x<0$ . The second rule applies because it uses $x\ge0$ . Check < versus <= before substituting.

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The definition chooses exactly one rule for each input, including the boundary x=0.

3. Preserve rational-expression exclusions

A rational expression is a fraction whose numerator and denominator are polynomials.

Return to

g(x)=\frac{x^2-1}{x^2-4x+3}.

Factor the denominator first:

x^2-4x+3=(x-1)(x-3).

The original function excludes

x=1\quad\text{and}\quad x=3.

Now factor the numerator and simplify:

\begin{aligned} g(x) &=\frac{(x-1)(x+1)}{(x-1)(x-3)}\\ &=\frac{x+1}{x-3}, \qquad x\ne1,\;x\ne3. \end{aligned}

Canceling $(x-1)$ does not make $x=1$ valid. The simplified formula agrees with the original only on the original domain.

Try it yourself:

Toggle the simplified function $s$ off and on, then compare g(1) with s(1). The curves overlap wherever the original function is defined, but only the simplified formula returns a value at $x=1$ .

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The curves overlap. Compare g(1), which is undefined, with s(1), which equals -1.

Feature

Hole

What creates it?

Every copy of a denominator factor cancels, leaving a finite simplified value at the excluded input.

In this example

$x=1$ , at the missing point $(1,-1)$

Feature

Vertical asymptote

What creates it?

A denominator factor remains after complete cancellation.

In this example

$x=3$

Feature

$x$ -intercept

What creates it?

The numerator is $0$ while the denominator is not $0$ .

In this example

$x=-1$

Feature	What creates it?	In this example
Hole	Every copy of a denominator factor cancels, leaving a finite simplified value at the excluded input.	$x=1$ , at the missing point $(1,-1)$
Vertical asymptote	A denominator factor remains after complete cancellation.	$x=3$
$x$ -intercept	The numerator is $0$ while the denominator is not $0$ .	$x=-1$

A zero from the numerator counts as an $x$ -intercept only when the original function is defined there.

Check your understanding:

For $f(x)=\frac{x^2-9}{x-3}$ , which statement is true?

Common mistake:

Restoring a canceled input. A canceled factor does not give permission to use the excluded input. Carry the original exclusions beside every simplified form.

4. Check candidates in the original equation

Consider

\sqrt{x+5}=x-1.

The square root is defined when $x\ge-5$ . It is also nonnegative, so equality requires

x-1\ge0,

which gives the stronger condition $x\ge1$ .

Squaring both sides gives

x+5=(x-1)^2,

which has the candidates $x=-1$ and $x=4$ . The value $-1$ already violates $x\ge1$ .

A candidate is a possible solution produced by a solving step that still needs to be checked. Check the remaining candidate in the original equation:

\sqrt9=3=4-1.

Only $x=4$ works.

An extraneous solution is a value produced by a transformed equation that does not satisfy the original equation.

Use the original equation as the final filter:

Record the original domain restrictions.
Solve or graph to find candidates.
Substitute each candidate into the original equation.
Keep only values that make every original expression defined and both sides equal.

This check matters after squaring, clearing denominators, taking even powers, or canceling factors.

Check your understanding:

The equation $\frac{x^2-4}{x-2}=\frac{16}{x+2}$ produces the candidates $x=2$ and $x=-6$ after simplification. Which is a solution to the original equation?

Common mistake:

Checking only the transformed equation. A candidate already satisfies the transformed equation that produced it. Substitute into the original equation, where the lost sign or domain condition still exists.

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Graph the original left and right sides. Their only intersection has x-coordinate 4.

5. Example: Choose the branch, then apply the condition

Worked example

The function $f$ is defined by

f(x)= \begin{cases} (x-2)^2, & x\ge2\\ 4-x, & x<2. \end{cases}

If $a>0$ , which choice gives a value of $a$ such that

f(a)=f(0)+5?

A
$-5$
B
$1$
C
$5$
D
$11$

Step 1

Evaluate the known input

Because $0<2$ , use $4-x$ :

f(0)=4.

Step 2

Find the target output

The condition becomes

f(a)=4+5=9.

Step 3

Graph the piecewise function and the target

Enter the compact definition and graph $y=9$ . The horizontal line meets the piecewise graph at $x=-5$ and $x=5$ .

The branch restrictions are already part of the function, so no invalid branch values appear.

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The graph enforces each branch condition. The prompt also requires a to be positive.

Step 4

Apply the remaining condition

The prompt requires $a>0$ , so reject $-5$ and keep

a=5.

The answer is C.

Common mistake:

Choosing $-5$ because it also gives the output $9$ . It satisfies the function equation but violates the stated condition $a>0$ .

Finish the solution

The original left and right sides are graphed. Read the visible intersection, then check the original exclusions before choosing an answer.

Keep only valid solutions

Finish the solution

Which value of $x$ satisfies

\frac{x^2-1}{x-1}=\frac4{x+1}?

First steps

Record the original exclusions: $x\ne1$ and $x\ne-1$ .
The two sides of the original equation are entered as separate graphs.
Select the visible intersection and record its $x$ -coordinate.
Compare that coordinate with the exclusions and answer choices.

Finish it

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Read the visible intersection of the original graphs, not unrestricted simplified formulas.

Practice problems

SAT practice problems

Keep the conditions visible. Record exclusions before simplifying, and check every final candidate against the original problem.

Use an included endpoint

Practice problem

The function

q(x)=2x+5

has domain

-2<x\le3.

What is the maximum value of $q$ on this domain?

Your answer

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Graph q with its full domain restriction, then inspect the included right endpoint.

Evaluate two branches

Practice problem

The function $f$ is defined by

f(x)= \begin{cases} 2x+5, & x<1\\ (x-1)^2, & x\ge1. \end{cases}

What is the value of

f(-2)+f(4)?

Your answer

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Define the piecewise function, then evaluate both requested inputs on a new line.

Classify rational-function features

Practice problem

The function $h$ is defined by

h(x)=\frac{(x+4)(x-2)}{(x-2)(x+1)}.

Which statement about the graph of $y=h(x)$ is true?

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Record the denominator zeros before canceling, then use the graph to confirm each feature.

Reject an extraneous radical solution

Practice problem

What is a solution to

\sqrt{m+9}=m-3?

Your answer

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Graph the original left and right sides, or solve algebraically and test every candidate.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Graph only the allowed inputs, and read a piecewise condition before choosing its rule.
Record every original denominator zero before canceling factors.
When every copy of a denominator factor cancels, its excluded input becomes a hole if the simplified value is finite. A factor that remains in the denominator creates a vertical asymptote.
A numerator zero is an $x$ -intercept only when the original function is defined there.
After squaring, clearing denominators, or canceling factors, test candidates in the original equation.
Use Desmos when it reveals the valid graph quickly, and use short algebra when it identifies an exact restriction faster.

Next lesson

Circle equations in the graph

Turn circle equations into visible centers, radii, intercepts, and geometric relationships.

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Free Practice

What you’ll learn

Prerequisites

The equation is not the whole story

Fast Desmos solution

Choose the right approach

Use a Desmos restriction when…

Write the restrictions first when…

1. Restrict a graph with curly braces

2. Enter a piecewise function

3. Preserve rational-expression exclusions

4. Check candidates in the original equation

5. Example: Choose the branch, then apply the condition

Evaluate the known input

Find the target output

Graph the piecewise function and the target

Apply the remaining condition

Finish the solution

Keep only valid solutions

First steps

Finish it

Practice problems

SAT practice problems

Use an included endpoint

Evaluate two branches

Classify rational-function features

Reject an extraneous radical solution

5 practice examples left

Quick recap

Circle equations in the graph

Free question bank practice problems