Exercise 7 Page 216

We want to graph the given polynomial function.

f (x) = (x - 2)^{2} (x + 1)

Let's start by finding the zeros of the function.

Zeros of the Function

To do so, we need to find the values of

x

for which

f (x) = 0 .

f (x) = 0 \Leftrightarrow (x - 2)^{2} (x + 1) = 0

Since the function is already written in factored form, we will use the Zero Product Property.

(x - 2)^{2} (x + 1) = 0

ZeroProdProp

Use the Zero Product Property

(x - 2)^{2} = 0 x + 1 = 0 (I) (II)

SqrtEqn

$(I):$ $LHS = RHS$

x - 2 = 0 x + 1 = 0

AddEqn

$(I):$ $LHS + 2 = RHS + 2$

x = 2 x + 1 = 0

SubEqn

$(II):$ $LHS - 1 = RHS - 1$

x = 2 x = - 1

We found that the zeros of the function are

- 1,

and

2 .

Graph

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$	$(x - 2)^{2} (x + 1)$	$f (x) = (x - 2)^{2} (x + 1)$
$- 0.5$	$(- 0.5 - 2)^{2} (- 0.5 + 1)$	$3.125$
$0$	$(0 - 2)^{2} (0 + 1)$	$4$
$1$	$(1 - 2)^{2} (1 + 1)$	$2$
$3$	$(3 - 2)^{2} (3 + 1)$	$4$

The points

(- 0.5, 3.125),

(0, 4),

(1, 2),

and

(3, 4)

are on the graph of the function. Finally, let's apply the Distributive Property. This will simplify the equation and determine the leading coefficient and degree of the polynomial function.

f (x) = (x - 2)^{2} (x + 1)

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

f (x) = (x^{2} - 2 x (2) + 2^{2}) (x + 1)

CalcPowProd

Calculate power and product

f (x) = (x^{2} - 4 x + 4) (x + 1)

▼

Simplify right-hand side

Distr

Distribute $(x + 1)$

f (x) = x^{2} (x + 1) - 4 x (x + 1) + 4 (x + 1)

Distr

Distribute $x^{2}$

f (x) = x^{3} + x^{2} - 4 x (x + 1) + 4 (x + 1)

Distr

Distribute $- 4 x$

f (x) = x^{3} + x^{2} - 4 x^{2} - 4 x + 4 (x + 1)

Distr

Distribute $4$

f (x) = x^{3} + x^{2} - 4 x^{2} - 4 x + 4 x + 4

AddSubTerms

Add and subtract terms

f (x) = x^{3} - 3 x^{2} + 4

We can see now that the leading coefficient is

1,

which is a positive number. Also, the degree is

3,

which is an odd number. Therefore, the end behavior is down and up.