Determining the Rule for the Graph of a Polynomial Revision Notes for VCE SSCE Mathematical Methods

Determining the Rule for the Graph of a Polynomial

Introduction

When working with polynomial functions, we often need to determine the rule (equation) from information shown on a graph. The key principle to remember is this: a polynomial function of degree $n$ is completely determined by any $n + 1$ points on the curve.

infoNote

This principle applies to all polynomial functions:

A straight line (degree 1) needs 2 points
A quadratic function (degree 2) needs 3 points
A cubic function (degree 3) needs 4 points
A quartic function (degree 4) needs 5 points

This means that if we know enough points on a polynomial graph, we can work out its exact equation.

Forms of cubic functions

Cubic functions can be written in several different forms, and recognizing which form to use depends on the information given in the graph.

Translated form

If you can identify the point of inflection $(h, k)$ , the cubic may have the form:

f(x) = a(x - h)^3 + k

In this case, you need one additional point to find the value of $a$ .

Factored form with three distinct roots

If the graph crosses the $x$ -axis at three different points, say $x = \alpha$ , $x = \beta$ , and $x = \gamma$ , then:

y = a(x - \alpha)(x - \beta)(x - \gamma)

You need one more point (often the $y$ -intercept) to find $a$ .

Factored form with a repeated root

If the graph touches the $x$ -axis at one point and crosses at another, there's a repeated factor. For example, if it touches at $x = \alpha$ and crosses at $x = \beta$ :

y = a(x - \alpha)^2(x - \beta)

Again, one additional point determines $a$ .

General form

When no special features are obvious, use the general cubic form:

y = ax^3 + bx^2 + cx + d

You'll need four points to create simultaneous equations for $a$ , $b$ , $c$ , and $d$ .

chatImportant

Choosing the Right Form:

Always examine the graph carefully before choosing which form to use:

Can you identify a point of inflection? → Use translated form
Does it cross the $x$ -axis three times? → Use factored form with distinct roots
Does it touch the $x$ -axis once and cross once? → Use repeated factor form
No clear pattern? → Use general form

Worked example: Finding the coefficient $a$

lightbulbExample

Worked Example: Using the Translated Form

Problem: A cubic function has rule of the form $y = a(x - 2)^3 + 2$ . The point $(3, 10)$ is on the graph. Find the value of $a$ .

Working:

The point of inflection is at $(2, 2)$ , so we use the form $y = a(x - 2)^3 + 2$ .

Substitute the point $(3, 10)$ :

10 = a(3 - 2)^3 + 2

10 = a(1)^3 + 2

10 = a + 2

8 = a

Therefore, $a = 8$ .

Explanation: When the coordinates of the point of inflection are known, and we have one other point on the graph, we can substitute directly to find $a$ .

lightbulbExample

Worked Example: Using Three x-axis Intercepts

Problem: A cubic function has rule of the form $y = a(x - 1)(x + 2)(x - 4)$ . The point $(5, 16)$ is on the graph. Find the value of $a$ .

Working:

The $x$ -axis intercepts are at $x = 1$ , $x = -2$ , and $x = 4$ .

Substitute the point $(5, 16)$ :

16 = a(5 - 1)(5 + 2)(5 - 4)

16 = a(4)(7)(1)

16 = 28a

a = \frac{16}{28} = \frac{4}{7}

Therefore, $a = \frac{4}{7}$ .

Explanation: When three $x$ -axis intercepts are known, these give us the three linear factors. We need one additional point to find the coefficient $a$ .

lightbulbExample

Worked Example: Finding Two Unknowns

Problem: A cubic function has rule of the form $f(x) = ax^3 + bx$ . The points $(1, 16)$ and $(2, 30)$ are on the graph. Find the values of $a$ and $b$ .

Working:

We know $f(1) = 16$ and $f(2) = 30$ .

Substituting $(1, 16)$ :

16 = a(1)^3 + b(1)

16 = a + b \text{ ... (1)}

Substituting $(2, 30)$ :

30 = a(2)^3 + b(2)

30 = 8a + 2b \text{ ... (2)}

Multiply equation (1) by 2:

32 = 2a + 2b \text{ ... (3)}

Subtract equation (3) from equation (2):

30 - 32 = 8a - 2a

-2 = 6a

a = -\frac{1}{3}

Substitute into equation (1):

16 = -\frac{1}{3} + b

b = 16 + \frac{1}{3} = \frac{49}{3}

Therefore, $a = -\frac{1}{3}$ and $b = \frac{49}{3}$ .

Explanation: When we have two unknowns, we need two points to create simultaneous equations, which we then solve.

Worked example: Finding all four coefficients

lightbulbExample

Worked Example: Finding All Four Coefficients Using Simultaneous Equations

Problem: For the cubic function with rule $f(x) = ax^3 + bx^2 + cx + d$ , the points $(-1, -18)$ , $(0, -5)$ , $(1, -4)$ and $(2, -9)$ lie on the graph. Find the values of $a$ , $b$ , $c$ and $d$ .

Working:

Substitute each point into the general form:

Point $(-1, -18)$ :

-18 = a(-1)^3 + b(-1)^2 + c(-1) + d

-a + b - c + d = -18 \text{ ... (1)}

Point $(0, -5)$ :

-5 = a(0)^3 + b(0)^2 + c(0) + d

d = -5 \text{ ... (2)}

Point $(1, -4)$ :

-4 = a(1)^3 + b(1)^2 + c(1) + d

a + b + c + d = -4 \text{ ... (3)}

Point $(2, -9)$ :

-9 = a(2)^3 + b(2)^2 + c(2) + d

8a + 4b + 2c + d = -9 \text{ ... (4)}

From equation (2), we know $d = -5$ .

Add equations (1) and (3):

2b + 2d = -22

2b + 2(-5) = -22

2b - 10 = -22

2b = -12

b = -6

Substitute $d = -5$ and $b = -6$ into equation (3):

a + (-6) + c + (-5) = -4

a + c = 7 \text{ ... (3')}

Substitute $d = -5$ and $b = -6$ into equation (4):

8a + 4(-6) + 2c + (-5) = -9

8a - 24 + 2c - 5 = -9

8a + 2c = 20 \text{ ... (4')}

Multiply equation (3') by 2:

2a + 2c = 14 \text{ ... (5)}

Subtract equation (5) from equation (4'):

8a - 2a = 20 - 14

6a = 6

a = 1

Substitute into equation (3'):

1 + c = 7

c = 6

Therefore, $a = 1$ , $b = -6$ , $c = 6$ and $d = -5$ .

chatImportant

Exam Tip: You can check your answer by verifying that all four points satisfy the equation $f(x) = x^3 - 6x^2 + 6x - 5$ .

Worked example: Using x-axis intercepts

lightbulbExample

Worked Example: Using Three Distinct X-axis Intercepts

Problem: The graph shown is that of a cubic function. Find the rule for this cubic function.

Working:

From the graph, the $x$ -axis intercepts are at $x = -3$ , $x = 1$ and $x = 4$ .

Therefore, the function has the form:

y = a(x - 4)(x - 1)(x + 3)

The graph passes through the point $(0, 4)$ . Substitute this point:

4 = a(0 - 4)(0 - 1)(0 + 3)

4 = a(-4)(-1)(3)

4 = 12a

a = \frac{1}{3}

The rule is $y = \frac{1}{3}(x - 4)(x - 1)(x + 3)$ .

Explanation: When the graph crosses the $x$ -axis at three distinct points, each intercept gives a linear factor. The $y$ -intercept (where $x = 0$ ) is then used to find the value of $a$ .

Worked example: Using a repeated factor

lightbulbExample

Worked Example: Recognizing and Using a Repeated Factor

Problem: The graph shown is that of a cubic function. Find the rule for this cubic function.

Working:

From the graph, we can see that:

The graph touches the $x$ -axis at $x = -3$ (it doesn't cross)
The graph crosses the $x$ -axis at $x = 1$

Because the graph touches at $x = -3$ , this is a repeated factor. Therefore, the function has the form:

y = k(x - 1)(x + 3)^2

The graph passes through the point $(0, 9)$ . Substitute this point:

9 = k(0 - 1)(0 + 3)^2

9 = k(-1)(9)

9 = -9k

k = -1

The rule is $y = -(x - 1)(x + 3)^2$ .

Explanation: When a graph touches the $x$ -axis at a point (rather than crossing it), that $x$ -value is a repeated root. For a cubic, this means the factor appears twice (squared). This is different from a simple crossing, which gives a single linear factor.

chatImportant

Exam Tip: Always check whether the graph crosses or touches the $x$ -axis at each intercept. Touching indicates a repeated factor.

Remember: "Touch means twice" — if a graph touches the $x$ -axis, that factor appears squared.

Worked example: Using four general points

lightbulbExample

Worked Example: Using Four General Points (No X-intercepts Given)

Problem: The graph of a cubic function passes through the points $(0, 1)$ , $(1, 4)$ , $(2, 17)$ and $(-1, 2)$ . Find the rule for this cubic function.

Working:

The cubic function has the general form:

y = ax^3 + bx^2 + cx + d

From the point $(0, 1)$ :

1 = a(0)^3 + b(0)^2 + c(0) + d

d = 1

Using the point $(1, 4)$ :

4 = a + b + c + 1

a + b + c = 3 \text{ ... (1)}

Using the point $(2, 17)$ :

17 = 8a + 4b + 2c + 1

8a + 4b + 2c = 16 \text{ ... (2)}

Using the point $(-1, 2)$ :

2 = -a + b - c + 1

-a + b - c = 1 \text{ ... (3)}

Add equations (1) and (3):

2b = 4

b = 2

Substitute $b = 2$ into equations (1) and (2):

a + c = 1 \text{ ... (4)}

8a + 2c = 8 \text{ ... (5)}

Multiply equation (4) by 2:

2a + 2c = 2 \text{ ... (6)}

Subtract equation (6) from equation (5):

6a = 6

a = 1

From equation (4):

1 + c = 1

c = 0

The rule is $y = x^3 + 2x^2 + 1$ .

Explanation: When no $x$ -intercepts or special features are given, use the general form. Four points give four equations, which can be solved systematically to find all four coefficients.

Using technology

infoNote

Modern calculators can solve systems of simultaneous equations quickly:

TI-Nspire: Use menu > Algebra > Solve System of Equations, or use the solve command with 'and' between equations.

Casio ClassPad: Define the function, then use the simultaneous equations icon to enter and solve the system.

While technology is helpful for checking, you should understand the manual method for exams.

Summary

bookmarkSummary

Key Points to Remember:

A polynomial of degree $n$ requires $n + 1$ points to determine its equation completely
For cubic functions, identify which form is most appropriate based on the given information
When a graph crosses the $x$ -axis, that gives a simple linear factor
When a graph touches the $x$ -axis, that indicates a repeated factor (appears squared)
The $y$ -intercept (point where $x = 0$ ) is often the easiest additional point to use
Always check your answer by substituting points back into your final equation

Determining the Rule for the Graph of a Polynomial (VCE SSCE Mathematical Methods): Revision Notes