When plotting polynomials the key points to look for are the x-intercepts and y-intercepts, as well as the behaviour of the graph as x goes to -\infty and \infty.

As with any graph, the x-intercepts will be the points on the graph where y=0 and the y-intercept will be the point on the graph where x=0. To find the intercepts it can help to first factorise the polynomial. Generalising from parabolas , the x-intercepts will be the zeros of the polynomial. That is, if a polynomial can be written as y=a(x-p)(x-q)(x-r)... then the x-intercepts will be at x=p,\,q,\,r,\,.... Further, the y-intercept will be at x=apqr.... Unlike quadratics, there is not necessarily a turning point at the midpoint of the x-intercepts.

The multiplicity of the zero also matters. The multiplicity is the number of times the factor goes into the polynomial. For example, in y=(x+2)^{2}(x-3),\,-2 has multiplicity 2 and 3 has multiplicity 1. If a zero has multiplicity 2 then the graph will touch the x-axis but not cross it. If a zero has multiplicity 3 then the graph will have a point of inflection, which means that the graph is horizontal at the point where it crosses the x-axis.

The behaviour of the graph as x goes to -\infty and \infty depends on the leading term of the polynomial. If the degree of the polynomial is odd it will increase on one end and decrease on the other. If it is even it will either increase on both ends or decrease on both ends. If the coefficient is positive the polynomial will increase as x increases and if it is negative it will decrease as x increases.

If the degree of the polynomial is odd:
- If the coefficient of the leading term is positive, the y-values will decrease as x goes to -\infty and increase as x goes to \infty.
- If the coefficient of the leading term is negative, the y-values will increase as x goes to -\infty and decrease as x goes to \infty.
If the degree of the polynomial is even:
- If the coefficient of the leading term is positive, the y-values will increase as x goes to -\infty and increase as x goes to \infty.
- If the coefficient of the leading term is negative, the y-values will decrease x goes to -\infty and decrease as x goes to \infty.

Examples

Example 1

The graph of the function y=f(x) is shown below.

-4

-3

-2

-1

-8

-6

-4

-2

What are the x-values of the x-intercepts of the graph?

Worked Solution

Apply the idea

The x-values of the x-intercepts are x=1,\,-2.

What is the y-value of the y-intercept of the graph?

Worked Solution

Apply the idea

The y-value of the y-intercept is y=-2.

What is the behaviour of the function as x \to \infty?

Worked Solution

Create a strategy

Look at the change in the value of y on the graph as we move from left to right.

Apply the idea

We can see that the graph is going down as we move to the right, so the function is decreasing.

What is the behaviour of the function as x \to -\infty?

Worked Solution

Create a strategy

Look at the change in the value of y on the graph as we move from right to left.

Apply the idea

We can see that the graph is going up as we move to the left, so the function is increasing.

Example 2

Consider the cubic function f(x)=4x^{3}+8x^{2}.

Complete the following table of values.

x	-6	-2	-1	0	1	2
f(x)

Worked Solution

Create a strategy

Substitute each x-value into the cubic function.

Apply the idea

For x=-6:

\displaystyle f(x)	\displaystyle =	\displaystyle 4x^{3}+8x^{2}	Write the function
	\displaystyle =	\displaystyle 4(-6)^{3}+8(-6)^{2}	Substitute x=-6
	\displaystyle =	\displaystyle -576	Evaluate

Similarly, by substituting the remaining x-values into f(x)=4x^{3}+8x^{2}, we get:

x	-6	-2	-1	0	1	2
f(x)	-576	0	4	0	12	64

Sketch the curve y=f(x).

Worked Solution

Create a strategy

Plot some of the points from the table and draw a curve through them.

Apply the idea

-4

-3

-2

-1

-12

-10

-8

-6

-4

-2

By plotting the points (-2,0),\,(-1,4),\,(0,0),\,(1,12) on a cartesian plane, we can sketch a curve through these points to draw the polynomial as shown.

Choose the four correct statements.

The curve has x-intercepts at -2 and 0.

At the point f(2), the function takes the value of 2.

The curve crosses the y-axis at the point x=0.

The value of the function is negative between x=-2 and x=0.

At the point x=2, the function takes the value of f(2).

As the value of x increases from 1 to infinity, the curve is increasing.

Worked Solution

Create a strategy

Examine the graph to determine which statements are correct.

Apply the idea

-4

-3

-2

-1

-12

-10

-8

-6

-4

-2

By looking at the graph, we can see that the curve touches (-2,0) and (0,0). So option A is correct.

At x=2, the y-value on the graph is not 2. It is a larger value than 12 since we can't see it on the graph. Also, we can look at the table in part (a) and at the point x=2, f(2)=64. So option B is not correct.

Same with option A, we can see that the curve goes through (0,0). So option C is correct.

Between x=-2 and x=0, the function is above the x-axis, so the function is positive. So option D is not correct.

When x=2, the y-value will equal f(2), so option E is correct.

The curve is moving upwards as x increases from 0. So option F is correct.

Example 3

A polynomial has been graphed with each of its intercepts shown.

-5

-4

-3

-2

-1

-6

-4

-2

Which of the following polynomials could describe the graph shown? (Note, k is a non-zero real number.)

y = k \left(x + 2\right)^{2} \left(x - 1\right) \left(x - 4\right)^{2}

y = k \left(x + 2\right) \left(x - 1\right)^{3} \left(x - 4\right)

y = k \left(x - 2\right)^{2} \left(x + 1\right) \left(x + 4\right)^{2}

y = k \left(x - 2\right)^{2} \left(x + 1\right)^{3} \left(x + 4\right)^{2}

Worked Solution

Create a strategy

Consider the x-intercepts and the shape of the curve at each intercept.

Apply the idea

The intercept at x=-4 corresponds to a factor of (x+4). Since the curve touches the x-axis at this point and bounces back, the multiplicity of this factor much be even.

This means that options A and B are not correct.

The intercept at x=-1 corresponds to a factor of (x+1). Since the curve crosses the x-axis at this point and looks like a straight line, the multiplicity must be 1.

This means that option D is not correct.

The correct option is C: y = k \left(x - 2\right)^{2} \left(x + 1\right) \left(x + 4\right)^{2}.

Use the y-intercept and the form of the polynomial from part (a) to determine the value of k.

Worked Solution

Create a strategy

Substitute the x and y values of the y-intercept into the equation from part (a).

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle k \left(x - 2\right)^{2} \left(x + 1\right) \left(x + 4\right)^{2}	Write the equation
\displaystyle 4	\displaystyle =	\displaystyle k(0-2)^{2}(0+1)(0+4)^{2}	Substitute x=0,\,y=4
\displaystyle 4	\displaystyle =	\displaystyle 64k	Evaluate
\displaystyle k	\displaystyle =	\displaystyle \dfrac{1}{16}	Divide both sides by 64

State the equation of the least degree polynomial that could have the graph displayed.

Worked Solution

Create a strategy

Substitute the value of k into the equation from part (a).

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle k \left(x - 2\right)^{2} \left(x + 1\right) \left(x + 4\right)^{2}	Write the equation
\displaystyle y	\displaystyle =	\displaystyle \dfrac{1}{16}\left(x - 2\right)^{2} \left(x + 1\right) \left(x + 4\right)^{2}	Substitute k=\dfrac{1}{16}

Idea summary

The x-intercepts of the graph of a polynomial are at the zeros of the polynomial. The y-intercept will be at the product of the zeros and the constant coefficient.

We can find these either by substituting y=0 and x=0 or by factorising the polynomial.

The multiplicity of the zero determines the behavious at the x-intercept:

Multiplicity 1 means the graph crosses the x-axis.
Multiplicity 2 means the graph touches the x-axis at a turning point.
Multiplicity 3 means the graph touches the x-axis at a point of inflection, which means that the graph will be horizontal at the x-intercept.

The behaviour of the graph as x goes to -\infty and \infty depends on the leading term of the polynomial:

If the degree of the polynomial is odd:
- If the coefficient of the leading term is positive, the y-values will decrease as x goes to -\infty and increase as x goes to \infty.
- If the coefficient of the leading term is negative, the y-values will increase as x goes to -\infty and decrease as x goes to \infty.
If the degree of the polynomial is even:
- If the coefficient of the leading term is positive, the y-values will increase as x goes to -\infty and increase as x goes to \infty.
- If the coefficient of the leading term is negative, the y-values will decrease x goes to -\infty and decrease as x goes to \infty.

Outcomes

MA5.3-9NA

sketches and interprets a variety of non-linear relationships

MA5.3-10NA

recognises, describes and sketches polynomials, and applies the factor and remainder theorems to solve problems

6.10 Sketching polynomials

Ideas

Graph of polynomials

Examples

Example 1

Example 2

Example 3

Outcomes

MA5.3-9NA

MA5.3-10NA

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