C3.5 Connect rational-function roots with intercepts

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Connecting algebraic solutions to points on a graph

A rational function is a function written as one polynomial divided by another. Its roots connect algebra to points where its graph meets the horizontal axis. One check matters: an input that makes the denominator zero is not in the function’s domain, so it cannot be a root or an intercept. This lesson reviews the needed ideas, shows the connection with an example and a table, and works through a full solution.

What you will learn

Explain how a rational-function root relates to an x-intercept.
Find x-intercepts by checking where the function equals zero and is defined.
Distinguish a numerator zero that gives an x-intercept from an input excluded from the domain.
Identify the y-intercept when it exists.

Prerequisite bridge: roots, intercepts, and domain

A root is an input value that makes a function’s output equal to zero. To find roots, solve the equation formed by setting the function equal to zero.

An x-intercept is a point where a graph meets the horizontal axis. Every point on that axis has a vertical coordinate of zero. Therefore, an x-intercept has the form

(a,0)

, where

a

is a root of the function.

The domain is the set of allowed input values. For a rational function, the denominator cannot equal zero. Any input that makes the denominator zero is excluded, even if an algebraic expression can be simplified.

The y-intercept is where the graph meets the vertical axis. Its input is zero, so it exists only when zero is in the domain. When it exists, its coordinates are

(0,f(0))

.

A function has an x-intercept at input

a

exactly when its value there is zero:

f(a)=0\Longleftrightarrow(a,0)

is an x-intercept.

A root is an input; an x-intercept is a point.
A rational function is undefined wherever its denominator is zero.
The y-intercept is found by evaluating the function at zero, if allowed.

Plain-language rule for rational functions

For a rational function, start by looking for inputs that make the numerator zero. A fraction equals zero when its numerator is zero and its denominator is not zero. So numerator zeros are candidates for roots, not automatic roots.

Check each candidate in the original denominator. If the denominator is nonzero there, the function’s value is zero, and the graph has an x-intercept at that input. If the denominator is zero there, the function is undefined at that input, so there is no root or x-intercept there.

This check still matters if a common factor can be cancelled. Cancelling a factor can give a simpler expression for inputs where the original function is defined. It does not put an excluded input back into the original domain. Keep the original denominator in mind when deciding whether a candidate is allowed.

The same domain check applies to the y-intercept. Substitute zero into the original function. If the original denominator is zero, there is no y-intercept, even if a simplified expression appears to have a value at zero.

For numerator

N(x)

and denominator

D(x)

, the condition for an x-intercept at

a

is

N(a)=0

and

D(a)\ne0\Longleftrightarrow(a,0)

is an x-intercept.

Set the numerator equal to zero to find possible x-intercept inputs.
Test each possible input in the original denominator.
A valid x-intercept is written as a point with vertical coordinate zero.

Multiple representations: equation, table, and graph

Consider the rational function

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

. The numerator is zero at

x=2

, while the denominator is zero at

x=-1

. The candidate

x=2

is allowed because the denominator is not zero there. Thus the graph has an x-intercept at

(2,0)

.

The table below makes the test visible. The row for

x=2

has output zero, so it gives the x-intercept. The row for

x=-1

has no output because the function is undefined there. It cannot give an intercept.

On a graph, the x-intercept is the point where the curve meets the horizontal axis. The algebra tells us where to look: solve the numerator equation, then confirm the input is allowed. The graph shows the same result as a point on the axis.

For the y-intercept of this function, use input zero. The output is

-2

, so the y-intercept is

(0,-2)

. This is a separate check: the y-intercept does not come from solving for a numerator zero.

The equation identifies candidates.
The table displays the function’s value or shows that it is undefined.
The graph represents a valid root as an x-intercept.

Guided example and application

In the worked example, the numerator has two factors that make it zero. One candidate is also excluded by the denominator. Comparing the candidates with the original denominator shows why a numerator zero alone is not enough.

For any new rational function, use the same reasoning. First identify the numerator’s zeros. Next reject any value that makes the original denominator zero. Finally, write each remaining x-intercept as a point. To find a y-intercept, check whether input zero is allowed and evaluate the function there.

This method connects the algebraic meaning of a root with the graph meaning of an intercept. The equation provides candidates, and the domain check confirms which candidates belong to the function.

Keep track of excluded inputs from the original denominator.
Report x-intercepts as ordered pairs.
Check the y-intercept separately by testing input zero.

Candidate inputs for $h(x)$

Input	Numerator	Denominator	Conclusion
$-2$	Zero	Zero	Excluded; not a root or intercept
$3$	Zero	Nonzero	Root; x-intercept $(3,0)$
$0$	Nonzero	Nonzero	Allowed; y-intercept $(0,-3)$

Worked example

Finding the valid x-intercepts

For

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

, find the roots and x-intercepts. Also find the y-intercept, if it exists.

Record excluded inputs
The original denominator is zero at $x=-2$ and $x=1$ . Neither value belongs to the domain, so neither can be a root or an intercept input.
$x\ne-2,\quad x\ne1$
Find numerator candidates
The numerator is zero when either factor is zero. These values are candidates, but each must still be checked against the excluded inputs. (x+2)(x-3)=0\Longrightarrow x=-2 or x=3
Check the candidates
The candidate $x=-2$ is excluded by the original denominator, so it is not a root. At $x=3$ , the denominator is nonzero and the numerator is zero, so $x=3$ is a root and gives an x-intercept.
$h(3)=0\Longrightarrow(3,0)$
Check the y-intercept
Zero is not an excluded input. Substituting zero gives output $-3$ , so the graph has a y-intercept at $(0,-3)$ .
$h(0)=\frac{(-2)(-3)}{(2)(-1)}=-3$

Answer: The only root is

x=3

, and the x-intercept is

(3,0)

. The y-intercept is (0,-3).

Check: At

x=3

, the numerator is zero and the denominator is nonzero. At

x=-2

, the original denominator is zero, so that input is not in the domain. Substitution at zero gives

-3

.

Common mistakes and how to avoid them

Treating every zero of the numerator as a root.

Correction: Check the original denominator at each candidate. A numerator zero that is excluded from the domain is not a root.

Cancelling a common factor and then treating its zero as part of the domain.

Correction: A cancelled factor does not change which inputs were excluded from the original function. Keep those exclusions when finding roots and intercepts.

Writing a root as though it were an intercept point.

Correction: A root is an input value such as

x=3

. The matching x-intercept is the point

(3,0)

.

Assuming every rational function has a y-intercept.

Correction: Test input zero in the original denominator. If it is zero, the function has no y-intercept.

Lesson summary

A root is an input that makes a function’s value zero; its graph point is an x-intercept.
For a rational function, numerator zeros are candidates for roots.
Reject any candidate that makes the original denominator zero.
Find a y-intercept by evaluating at zero only when zero is in the domain.

Check your understanding

Question 1

For

f(x)=\frac{(x-4)(x+1)}{x+1}

, which value gives an x-intercept?

$(4,0)$
$(-1,0)$
$(0,4)$
There are no x-intercepts.

Show answer and explanation

(4,0)

The numerator candidates are

x=4

and

x=-1

. The original denominator excludes

x=-1

, but

x=4

is allowed and makes the function zero. The x-intercept is

(4,0)

.

Question 2

For

p(x)=\frac{x+3}{x-2}

, what is the x-intercept?

$(2,0)$
$(-3,0)$
$(0,-3)$
There is no x-intercept.

Show answer and explanation

(-3,0)

The numerator is zero at

x=-3

, and the denominator is nonzero there. Therefore the x-intercept is

(-3,0)

.

Question 3

For

q(x)=\frac{x-5}{x}

, does the graph have a y-intercept?

Yes, at $(0,-5)$ .
Yes, at $(5,0)$ .
No, because input zero is excluded.
No, because the numerator has no zero.

Show answer and explanation

No, because input zero is excluded.

The denominator is zero at input zero. The function is undefined there, so it has no y-intercept.

Key terms

Rational function: A function written as a polynomial divided by another polynomial, where the denominator is not zero.
Root: An input value that makes the function’s output equal to zero.
Intercept: A point where a graph meets one of the coordinate axes.
Domain: The set of input values for which a function is defined.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C3.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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C3.5 · Connect rational-function roots with intercepts

Ontario Grade 12 Mathematics

Connecting algebraic solutions to points on a graph

What you will learn

Prerequisite bridge: roots, intercepts, and domain

Plain-language rule for rational functions

Multiple representations: equation, table, and graph

Guided example and application

Candidate inputs for $h(x)$

Finding the valid x-intercepts

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson and its review