1. Polynomials:
Graphing Polynomials.
By Mr Porter

2. Assumed Knowledge:
It is assumed that students can fully factorise polynomials.
Also, student should know how to
* factorise quadratic expression, ax2 + bx + c
* determine if a quadratic is positive or negative DEFINITE
(using the discriminate ∆ = b2 – 4ac ).
* Solve simple linear equations, mx + b = 0
Student should be familiar with the coordinate number plane.
Quadrant 1 Q1
Quadrant 2 Q2
Quadrant 3 Q3
Quadrant 4 Q4
X-axis
Y-axis
Note:
Y-axis is also the polynomial P(x).

3. Example 1:
Graph the factorised polynomial, P(x) = x ( x – 3 )( x + 2 ).
Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve.
x = 0
x – 3 = 0  x = 3
x +2 = 0  x = -2
Step 5: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = x ( x – 3 ) ( x + 2 )
Step 2: Draw a simple number plane, mark and label the three solution
x = -2 , x = 0 and x = 3.
P(x)
X-axis • 3 -2
Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x = 1x3
Note, it is POSITIVE.
We start in the first quadrant, Q1. x
These points are guesses! If you want accuracy, evaluate P(-1) and P(1.5)
Step 4: Find the vertical axis intercept (it is the constant term) Note x = (x – 0).
So, take each of the constant terms and multiply:
(0) x (-3) x (+2) = 0
The vertical intercept is 0.
Remember,
it is a sketch!

4. Example 2:
Graph the factorised polynomial, P(x) = x(x + 1)( x + 3 )( x – 4).
Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve.
x = 0
x + 1 = 0  x = -1
x + 3 = 0  x = -3
x – 4= 0  x = 4
Step 5: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = x(x + 1) ( x + 3 ) ( x – 4)
Step 2: Draw a simple number plane, mark and label the four solution
x = -3 , x = -1, x = 0 and x = 4.
P(x)
X-axis • 4 -3 -1
Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x x x = +1x4
Note, it is POSITIVE.
We start in the 1st quadrant, Q1. x
Step 4: Find the vertical axis intercept (it is the constant term) Note x = (x – 0).
So, take each of the constant terms and multiply:
(0) x (+1) x (+3) x (-4) = 0
The vertical intercept is 0.
Remember,
it is a sketch!

5. Example 3:
Graph the factorised polynomial, P(x) = (x + 1)( x + 3 )( 2 – x ).
Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve.
x + 1 = 0  x = -1
x + 3 = 0  x = -3
2 – x = 0  x = 2
Step 5: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = (x + 1) ( x + 3 ) ( 2 – x)
Step 2: Draw a simple number plane, mark and label the three solution
x = -3 , x = -1 and x = 2.
P(x)
X-axis • 2 -3 -1
Step 4: Find the vertical axis intercept (it is the constant term) So, take each of the constant terms and multiply:
(+1) x (+3) x (+2) = +6
The vertical intercept is +6.
6 •
Step 3: Starting point! Take each of the x-terms and multiply them. x x x x -x = -1x3
Note, it is NEGATIVE.
We start in the 4th quadrant, Q4. x
Remember,
it is a sketch!

6. Example 4:
Graph the factorised polynomial, P(x) = (x + 1)( x + 3 )( x – 4)(5 – x).
Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve.
x + 1 = 0  x = -1
x + 3 = 0  x = -3
x – 4 = 0  x = 4
5 – x = 0  x = 5
Step 5: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = (x + 1) ( x + 3 ) ( x – 4) (5 – x)
Step 2: Draw a simple number plane, mark and label the four solution
x = -3 , x = -1, x = 4 and x = 5.
P(x)
X-axis • 4 -3 -1 5
Remember,
it is a sketch!
Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x x -x = -1x4
Note, it is NEGATIVE.
We start in the 4th quadrant, Q4. x
Step 4: Find the vertical axis intercept (it is the constant term) So, take each of the constant terms and multiply:
(+1) x (+3) x (-4) x (+5)= -60
The vertical intercept is -60.
-60 •

7. Example 5: Sketch the polynomial, P(x) = x3 – 4x2 + 4x.
Step 5: Find the vertical axis intercept (it is the constant term) . In this case, it is the origin, (0, 0)
The vertical intercept is 0.
Step 6: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = x3 – 4x2 + 4x
Step 1: Factorise the polynomial.
P(x) = x3 – 4x2 + 4x
P(x) = x(x – 2) (x – 2)
Step 3: Draw a simple number plane, mark and label the two solution
x = 0 and x = 2.
P(x)
X-axis • 2
Step 2: Find the x-intercepts, take each factor and set them equal to zero and solve.
x = 0
x – 2 = 0  x = 2
Step 4: Starting point! Since the leading coefficient is +1 of x3
Note, it is POSITIVE.
We start in the 1st quadrant, Q1. x
Remember,
it is a sketch!
The two values at x = 2, means that the polynomial has a turning point at x = 2.

8. Example 6: Sketch the polynomial, P(x) = 2x3 + 7x2 + 4x – 4.
Step 5: Find the vertical axis intercept (it is the constant term) . In this case, it is -4
The vertical intercept is -4. • -4
Step 6: Start drawing the polynomial,
the curve must pass through each of the x-intercepts and the vertical intercept
P(x) = 2x3 + 7x2 + 4x – 4
Step 1: Factorise the polynomial.
P(x) = 2x3 + 7x2 + 4x – 4
P(x) = (2x – 1) (x + 2) (x + 2)
Step 3: Draw a simple number plane, mark and label the two solution
x = 1/2 and x = -2.
P(x)
X-axis • -2
1/2
Step 2: Find the x-intercepts, take each factor and set them equal to zero and solve.
2x – 1 = 0  x = 1/2
x + 2 = 0  x = -2
Step 4: Starting point! Since the leading coefficient is +2 of x3
Note, it is POSITIVE.
We start in the 1st quadrant, Q1. x
Remember,
it is a sketch!
The two values at x = -2, means that the polynomial has a turning point at x = -2.

9. Example Polynomials graphs generated using a FX-Graph application.
P(x) = x(x + 1) ( x + 3 ) ( x – 4)
P(x) = x ( x – 3 ) ( x + 2 )
Ex 1
Ex 2
P(x) = (x + 1) ( x + 3 ) ( 2 – x)
P(x) = (x + 1) ( x + 3 ) ( x – 4) (5 – x)
Ex 3
Ex 4