Exercise 15 Page 605

a Let's identify the values of

a,

b,

and

c

for the given function written in standard form.

h = - 16 t^{2} + 32 t ⇕ h = - 16 t^{2} + 32 t + 0

We see that

a = - 16,

b = 32,

and

c = 0 .

Since

a < 0,

the graph opens downward. Therefore, it has a maximum value.

b We know that the maximum value is the

y -

value of the vertex. The

x -

coordinate of the vertex is found using the following expression.

- \frac{b}{2 a}

We can find it since we have shown

a = - 16

and

b = 32

in Part A.

- \frac{b}{2 a}

SubstituteII

$a = - 16$ , $b = 32$

- \frac{3 2}{2 ( - 1 6 )}

▼

Evaluate

MultPosNeg

$a (- b) = - a \cdot b$

- \frac{3 2}{- 3 2}

RemoveNegFracAndDenom

$- \frac{a}{- b} = \frac{a}{b}$

\frac{3 2}{3 2}

CalcQuot

Calculate quotient

1

To find the

y -

coordinate, we need to substitute

1

for

t .

h = - 16 t^{2} + 32 t

Substitute

$t = 1$

h = - 16 (1)^{2} + 32 (1)

▼

Evaluate right-hand side

BaseOne

$1^{a} = 1$

h = - 16 (1) + 32 (1)

IdPropMult

Identity Property of Multiplication

h = - 16 + 32

AddTerms

Add terms

h = 16

The vertex is

(1, 16) .

Hence, the maximum

h -

value is

16 .

c To determine the domain and range, let's look at graph of the rocket's behavior.

Here, than range is the set of all the potential

h -

values. Notice that the graph goes no higher than

h = 16

and no lower than

h = 0 .

Therefore, the range is all real numbers between these two values, inclusively.

Range : {h ∣ 0 \leq h \leq 16}

It does not make sense for the rocket to go above it's maximum height

h = 16

or below ground level

h = 0 .

Similarly, we can think about the limits on the domain since

t

is measuring time. Note the events that occur at the following times.

t = 0 and t = 2

The given function models the path of the launched rocket from launch

t = 0

to landing

t = 2

making these the endpoints of the domain, inclusively.

Domain : {t ∣ 0 \leq t \leq 2}