Connect factored form with intercepts and sketches

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Use factors to locate and interpret a polynomial’s intercepts

Factored form makes some graph features easy to read. Each factor can show where a polynomial meets the x-axis. The exponent on that factor helps describe how the graph behaves at the intercept. Substituting zero gives the y-intercept. Together, these clues support a useful sketch without determining every point on the curve.

What you will learn

Find a polynomial’s x-intercepts from its factored form.
Use the exponent on a factor to tell whether the graph crosses or touches the x-axis at an intercept.
Find the y-intercept and use these features to sketch the graph.

1. Prerequisite bridge: intercepts and factors

An intercept is a point where a graph meets an axis. An x-intercept lies on the x-axis, so its y-coordinate is zero. A y-intercept lies on the y-axis, so its x-coordinate is zero.

A factor is an expression multiplied by another expression. For example, in

f(x)=(x-2)(x+1)

, the factors are

x-2

and

x+1

. A product equals zero if at least one factor equals zero. This is called the zero-product property.

To find x-intercepts from factored form, set each factor equal to zero. A factor of the form

x-a

gives the zero

x=a

, and therefore the intercept

(a,0)

. To find the y-intercept, substitute

x=0

and write the result as

(0,f(0))

.

f(a)=0

At an x-intercept, the function value is zero.
At a y-intercept, the input value is zero.
A factor $x-a$ gives the x-intercept $(a,0)$ .

2. What a factor’s exponent tells you

A zero is an input that makes the function value zero. Each real zero shown by a factor gives an x-intercept. For example, the factor

x-3

gives the zero

x=3

and the intercept

(3,0)

.

The exponent on a factor is called its multiplicity. At an intercept with odd multiplicity, the graph crosses the x-axis. At an intercept with even multiplicity, it touches the axis and turns back. The exponent tells you this local behaviour; it does not give the exact shape of the whole graph.

For example,

(x-a)^2

has an even exponent, so the graph touches and turns at

(a,0)

. The factor

(x-b)^3

has an odd exponent, so the graph crosses at

(b,0)

.

(x-a)^m=0

Odd multiplicity means the graph crosses the x-axis.
Even multiplicity means the graph touches the x-axis and turns.
The x-coordinate of the intercept comes from setting the factor equal to zero.

3. Connect the clues to a sketch

A sketch is a drawing that shows the important known features of a graph. Start by marking the x-intercepts from the factors. At each one, note whether the curve crosses or touches the axis. Then substitute

x=0

to find and mark the y-intercept.

If you want to check whether the graph is above or below the x-axis between intercepts, choose an input in that interval and evaluate the function. A positive value places the graph above the axis at that input; a negative value places it below.

Draw a smooth curve that fits the intercepts and the crossing or touching behaviour. These clues do not specify every point or the exact location of every bend, so do not treat a sketch as an exact plot.

Mark the x-intercepts and y-intercept before drawing the curve.
Use the factor exponents to show crossings and touches.
A test input can check whether the graph is above or below the axis at that point.

Factor exponent and behaviour at the x-intercept

Factor	Zero	Multiplicity	Graph behaviour
$(x-a)^1$	$x=a$	Odd	Crosses the axis
$(x-a)^2$	$x=a$	Even	Touches and turns
$(x-a)^3$	$x=a$	Odd	Crosses the axis
$(x-a)^4$	$x=a$	Even	Touches and turns

Worked example

Sketching from factored form

Sketch

f(x)=-(x+2)^2(x-1)

by identifying its intercepts and its behaviour at each x-intercept.

Find the x-intercepts
Set each distinct factor equal to zero. The factor $x+2$ gives $x=-2$ , and the factor $x-1$ gives $x=1$ . Write each as a point on the x-axis.
$(-2,0), (1,0)$
Read the behaviour at each intercept
The factor $x+2$ is squared, so its zero has even multiplicity. The graph touches the axis and turns at $(-2,0)$ . The factor $x-1$ has exponent one, so the graph crosses the axis at $(1,0)$ .
$(x+2)^2, (x-1)^1$
Find the y-intercept
Substitute $x=0$ . The function value is four, so the y-intercept is $(0,4)$ .
$f(0)=-(2)^2(-1)=4$
Check the graph’s position
Choose $x=-1$ , which lies between the two x-intercepts. The function value is positive, so the graph is above the x-axis at $x=-1$ . Use this point along with the intercepts to guide the curve.
$f(-1)=-(1)^2(-2)=2$

Answer: The sketch has x-intercepts

(-2,0)

and

(1,0)

, and a y-intercept

(0,4)

. It touches and turns at

x=-2

, crosses at

x=1

, and is above the x-axis at

x=-1

.

Check: The intercepts follow directly from the factors. Substituting

x=0

gives

4

, and substituting

x=-1

gives

2

, so both values are consistent with the stated points.

Common mistakes and how to avoid them

Reading the factor

x+2

as giving the zero

x=2

.

Correction: Set the factor equal to zero. Solving

x+2=0

gives

x=-2

, so the intercept is

(-2,0)

.

Assuming that every x-intercept is a crossing.

Correction: Check the exponent on the factor. Odd multiplicity means crossing; even multiplicity means touching and turning.

Using an x-intercept as the y-intercept.

Correction: For the y-intercept, substitute

x=0

. The point is

(0,f(0))

.

Drawing a curve that ignores one of the intercepts.

Correction: Mark all x-intercepts and the y-intercept first. Then draw a curve consistent with those points and each crossing or touch.

Lesson summary

A factor $x-a$ gives an x-intercept at $(a,0)$ .
Odd multiplicity means the graph crosses the x-axis; even multiplicity means it touches and turns.
Substitute $x=0$ to find the y-intercept.
Use the intercepts and crossing or touching behaviour to guide a sketch.
A test input can show whether the graph is above or below the x-axis at that input.

Check your understanding

Question 1

For

g(x)=(x-4)^2(x+1)

, what happens at the x-intercept where

x=4

?

The graph crosses the axis.
The graph touches the axis and turns.
There is no x-intercept at $x=4$ .
The point is a y-intercept.

Show answer and explanation

The graph touches the axis and turns.

The factor

(x-4)^2

gives the zero

x=4

with even multiplicity. The graph touches and turns at

(4,0)

.

Question 2

For

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

, what is the y-intercept?

$(0,-48)$
$(0,48)$
$(0,-12)$
$(0,12)$

Show answer and explanation

(0,-48)

Substitute

x=0

:

h(0)=2(3)(-2)^3=-48

. Therefore, the y-intercept is

(0,-48)

.

Question 3

For

p(x)=(x+2)(x-5)^2

, which statement correctly describes the x-intercepts?

The graph crosses at $(-2,0)$ and crosses at $(5,0)$ .
The graph touches at $(-2,0)$ and crosses at $(5,0)$ .
The graph crosses at $(-2,0)$ and touches at $(5,0)$ .
The graph touches at both intercepts.

Show answer and explanation

The graph crosses at

(-2,0)

and touches at

(5,0)

.

The factor

x+2

has exponent one, so the graph crosses at

(-2,0)

. The factor

(x-5)^2

has even multiplicity, so the graph touches and turns at

(5,0)

.

Key terms

Factor: An expression multiplied by another expression to form a product.
Zero: An input value that makes the function value equal to zero.
Multiplicity: The exponent on the factor that gives a zero.
Sketch: A drawing that shows important known features of a graph without claiming to plot every point exactly.

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