McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
Study Guide and Review
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Exercise 15 Page 605

Practice makes perfect
a

Let's identify the values of a, b, and c for the given function written in standard form.

h=- 16t^2+32t ⇕ h= - 16t^2+ 32t+ 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.

- b/2 aWe can find it since we have shown a= - 16 and b= 32 in Part A.

- b/2a
- 32/2( - 16)
â–¼
Evaluate
- 32/- 32
32/32
1

To find the y-coordinate, we need to substitute 1 for t.

h= -16t^2+32t
h= -16( 1)^2+32( 1)
â–¼
Evaluate right-hand side
h= -16(1)+32(1)
h= -16+32
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 ≤ h ≤ 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 andt=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 ≤ t ≤ 2 }