Graph and analyze polynomial functions up to degree five

Alberta Grade 12 Mathematics

Relations and Functions

Zeros, multiplicity, shape, and end behaviour for degrees one to five

A polynomial graph can be understood by combining a few clues: its degree, its leading coefficient, its zeros, and its intercepts. This lesson uses those clues to graph and analyze polynomial functions of degree five or less. You will also review how factors identify zeros and how a graph behaves near a zero. No calculus is needed.

What you will learn

Recognize a polynomial function and identify its degree and leading coefficient.
Use zeros, multiplicities, and intercepts to sketch and analyze a polynomial graph.
Describe a polynomial’s end behaviour and domain.
Check a graph using exact values and distinguish them from estimates.

1. Review: terms, factors, and zeros

A polynomial is an expression made from terms such as constants and powers of a variable, combined by addition or subtraction. In a polynomial, the exponents on the variable are non-negative whole numbers. For example,

2x^3-x+4

is a polynomial. An expression with

x

in a denominator or with a negative exponent is not a polynomial.

The degree is the greatest exponent with a non-zero coefficient. In

2x^3-x+4

, the degree is three. The leading coefficient is the coefficient of the term with the greatest exponent, so here it is

2

. These two features help predict how the graph behaves at its far left and right.

A zero is an input that makes the function’s value zero. On a graph, a real zero is an

x

-intercept. Factoring helps find zeros: if a polynomial has a factor

x-a

, then substituting

x=a

makes that factor zero. A factor written more than once gives a repeated zero, also called a zero with multiplicity greater than one.

If

x-a

is a factor, then substituting

a

makes the polynomial equal to zero.

The degree is the greatest exponent; the leading coefficient belongs to that term.
A real zero gives an $x$ -intercept.
A polynomial’s domain is all real numbers.

2. Predict the ends and trace the zeros

End behaviour describes what the graph does far to the left and far to the right. For an even degree, both ends point in the same direction. They point up when the leading coefficient is positive and down when it is negative. For an odd degree, the ends point in opposite directions. A positive leading coefficient gives a left end down and a right end up; a negative leading coefficient reverses those directions.

The multiplicity of a zero helps describe how the graph meets the

x

-axis there. At a zero with multiplicity one, the graph crosses the axis. At a zero with even multiplicity, it touches the axis and turns back. At a zero with odd multiplicity greater than one, it crosses but often looks flatter near the axis. These are useful shape clues, not a replacement for checking the function’s values.

The

y

-intercept is found by setting

x=0

. To sketch, mark the intercepts and zeros, indicate the end behaviour, then connect the clues with a smooth curve. A degree-

n

polynomial can have at most

n

real zeros and at most

n-1

turning points. A turning point is where the graph changes from rising to falling or from falling to rising. A sketch may show fewer; the degree gives an upper limit, not a promise.

n-1

Even degree: ends point the same way. Odd degree: ends point opposite ways.
A repeated factor helps determine whether the graph crosses or touches at its zero.
Use intercepts and end behaviour together; no single clue gives the whole graph.

3. Verify a sketch and describe the function

A graph is an estimate of a function’s values between known points. Substitution gives exact points that can test the sketch. For instance, if a proposed point has an integer input, substitute that input into the polynomial and compare the exact output with the graph.

When describing a polynomial graph, state its domain, range when it can be read from the graph, intercepts, and intervals where it rises or falls if these are requested. The domain of every polynomial is all real numbers. The range depends on the graph. For example, an even-degree graph with both ends up may have a lowest point, but its exact range should be stated only when that lowest value is known.

A graphing window shows only part of the coordinate plane. A curve may continue beyond the displayed edges. Do not treat the edge of the window as an endpoint of the function. If intercepts or turning points are hard to read, use the factored form and selected substitutions to check the picture.

y=f(x)

Substitution checks a plotted point exactly.
Every polynomial has domain all real numbers.
A graphing window is only a view of the graph, not its full domain.

Worked example

A cubic with three simple zeros

Analyze and sketch

f(x)=x^3-3x

. State its intercepts, end behaviour, and domain.

Cubic with three intercepts

The graph shows only a window of the full domain.

Factor to find the zeros
Factor out the common factor $x$ . The product is zero when any factor is zero, so the three zeros are $-\sqrt{3}$ , $0$ , and $\sqrt{3}$ . Each factor occurs once, so the graph crosses the axis at each zero.
$f(x)=x(x^2-3)=x(x-\sqrt{3})(x+\sqrt{3})$
Use degree and leading coefficient
The degree is three and the leading coefficient is positive. The degree is odd, so the left end points down and the right end points up. The $y$ -intercept is found by evaluating the function at zero; it is also the origin.
$f(0)=0$
State the domain and check a point
A polynomial is defined for every real input, so the domain is all real numbers. Substituting $x=1$ gives an exact check point. The graph should pass below the axis at that input.
$f(1)=1-3=-2$

Answer: The

x

-intercepts are

(-\sqrt{3},0)

,

(0,0)

, and

(\sqrt{3},0)

. The

y

-intercept is

(0,0)

. The graph crosses at all three zeros, falls to the left, and rises to the right. Its domain is all real numbers.

Check: The factorization expands to

x^3-3x

, confirming the zeros and the original function.

Worked example

A quartic with a repeated zero

Analyze

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

. Identify its zeros, intercepts, and end behaviour. Use one substitution to check the sketch.

Quartic with a repeated zero

The graph shows only a window of the full domain.

Read zeros from the factors
The factor $x^2$ gives a zero at zero with multiplicity two. The other factors give zeros at $-2$ and $1$ , each with multiplicity one. The graph touches and turns at zero, and crosses at $-2$ and $1$ .
$x^2(x+2)(x-1)=0$
Find degree and end behaviour
The degree is four because the factor degrees add to four. Its leading coefficient is positive. An even degree with a positive leading coefficient has both ends pointing up. The $y$ -intercept is zero because the function contains the factor $x^2$ .
$g(x)=x^4+x^3-2x^2$
Check a point
At $x=-1$ , the function is negative. This confirms that the graph lies below the axis between the zeros at $-2$ and zero. Together with the crossings and touch, this helps guide a consistent sketch.
$g(-1)=1-1-2=-2$

Answer: The

x

-intercepts are

(-2,0)

,

(0,0)

, and

(1,0)

. The graph crosses at

-2

and

1

, and touches and turns at zero. Both ends point up. The

y

-intercept is

(0,0)

, and the domain is all real numbers.

Check: The substitution gives

g(-1)=-2

, so the point lies below the axis. This agrees with the graph’s position between

-2

and zero.

Worked example

A fifth-degree function with two distinct zeros

Analyze

h(x)=(x+2)^2(x-1)^3

. Describe its zeros, end behaviour, intercepts, and domain.

Fifth-degree graph with repeated zeros

The graph shows only a window of the full domain.

Interpret the factors
The zero $-2$ has multiplicity two, so the graph touches and turns there. The zero $1$ has multiplicity three, so the graph crosses there and may appear flatter near the axis. The degree is five.
$(x+2)^2(x-1)^3=0$
Predict the ends
The leading term is $x^5$ , so the leading coefficient is positive and the degree is odd. The graph falls to the left and rises to the right. The $y$ -intercept is found by setting $x=0$ .
$h(0)=2^2(-1)^3=-4$
State the domain and intercepts
The $y$ -intercept is $(0,-4)$ . The zeros give the $x$ -intercepts $(-2,0)$ and $(1,0)$ . Since the function is a polynomial, every real number is allowed as an input.
$\text{domain}=\mathbb{R}$

Answer: The graph touches and turns at

x=-2

, crosses at

x=1

, falls to the left, and rises to the right. Its

y

-intercept is

(0,-4)

, and its domain is all real numbers.

Check: The multiplicities add to five, matching the degree. The positive leading coefficient and odd degree agree with the stated end behaviour.

Common mistakes and how to avoid them

Assuming every zero is a crossing.

Correction: Check the multiplicity. An even multiplicity gives a touch and turn; an odd multiplicity gives a crossing.

Using the degree alone to decide which way the ends point.

Correction: Use both the degree and the sign of the leading coefficient.

Treating the edge of a graphing window as the end of a polynomial’s domain.

Correction: A polynomial’s domain is all real numbers. The window only limits what is displayed.

Reporting an estimated graph coordinate as exact.

Correction: Use exact factor values or substitution when possible. Identify a read-from-graph value as an estimate.

Lesson summary

Degree and leading coefficient determine polynomial end behaviour.
Factors give zeros; multiplicity helps show whether the graph crosses or touches at each zero.
The $y$ -intercept is found by setting $x=0$ .
Every polynomial has domain all real numbers.
Use exact substitutions to check points and use a graph as a visual model.

Check your understanding

Question 1

A polynomial has degree four and a negative leading coefficient. What is its end behaviour?

Both ends point down.
The left end points down and the right end points up.
Both ends point up.
The left end points up and the right end points down.

Show answer and explanation

Both ends point down.

An even degree gives ends in the same direction. A negative leading coefficient makes both ends point down.

Question 2

For

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

, what happens at the zero

x=3

?

The graph crosses the axis.
The graph touches the axis and turns back.
The graph has no intercept there.
The graph has a vertical asymptote there.

Show answer and explanation

The graph touches the axis and turns back.

The factor

(x-3)

is repeated twice, so the zero has even multiplicity. The graph touches and turns at that zero.

Question 3

What is the

y

-intercept of

q(x)=x^3+2x^2-5

?

$(0,5)$
$(0,-5)$
$(5,0)$
$(-5,0)$

Show answer and explanation

(0,-5)

Set

x=0

. Then

q(0)=-5

, so the

y

-intercept is

(0,-5)

.

Key terms

Polynomial: An expression made from constants and non-negative whole-number powers of a variable, combined by addition or subtraction.
Degree: The greatest exponent in a polynomial with a non-zero coefficient.
Leading coefficient: The coefficient of the term with the greatest exponent.
Zero: An input that makes a function’s value zero.
Multiplicity: The number of times a factor occurs in a factored polynomial.
End behaviour: The directions a graph goes as the input moves far left or far right.
Turning point: A point where a graph changes from rising to falling or from falling to rising.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation RF12. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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RF12 · Graph and analyze polynomial functions up to degree five

Alberta Grade 12 Mathematics

Zeros, multiplicity, shape, and end behaviour for degrees one to five

What you will learn

1. Review: terms, factors, and zeros

2. Predict the ends and trace the zeros

3. Verify a sketch and describe the function

A cubic with three simple zeros

A quartic with a repeated zero

A fifth-degree function with two distinct zeros

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through Math 30-1

About this lesson and its review