Exercise 106 Page 406

Practice makes perfect

To find the roots of a function, we need to find the solutions when y=0.

0=x^3-9x To solve this polynomial equation, we will start by factoring out the greatest common factor (GCF).

0=x^3-9x

FactorOut

Factor out x

0=x(x^2-9)

RearrangeEqn

Rearrange equation

x(x^2-9)=0

We have rewritten the left-hand side as a product of two factors. Now, we will apply the Zero Product Property to solve the equation.

x(x^2-9) =0

ZeroProdProp

Use the Zero Product Property

lcx=0 & (I) x^2-9=0 & (II)

From Equation (I), we found that one solution is x=0. To find other solutions, we will solve Equation (II). Note that this is a quadratic equation. We will solve this equation by taking the square root of both sides of the equation.

x^2-9=0

AddEqn

LHS+9=RHS+9

x^2 = 9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x = ± sqrt(9)

CalcRoot

Calculate root

x = ± 3

These solutions to the quadratic equation are also solutions for the equation x^3-9x=0. Recall that x=0 is also the solution to the equation. Therefore, the roots of the given function are points (0,0), (3,0), and (-3,0).

Let's make a table of values to sketch a graph of the given function. When you are making a table of values make sure to use a variety of points, including negative and positive values. We can also use the roots of the function that we calculated in Part A.

x	x^3-9x	y=x^3-9x
- 3	( - 3)^3-9( - 3)	0
- 2	( - 2)^3-9( - 2)	10
- 1	( - 1)^3-9( - 1)	8
0	0^3-9( 0)	0
1	1^3-9( 1)	- 8
2	2^3-9( 2)	-10
3	3^3-9( 3)	0

We will now plot the obtained points and connect them with a smooth curve. Consider also that this is an odd-degree polynomial with a positive leading coefficient. &f(x) → -∞ as x → - ∞ &f(x) → + ∞ as x → + ∞ This tells us about the end behavior of the function.