Page 60 - Math Course 1 (Book 2)

Relations as a Function

B. Find the value h(2z).
Let’s Begin
h(2z) = 160(2z) + 16(2z) 2 Replace t with 2z.
= 320z + 64z 2 Multiply.

Function Values

Answer h(2z) = 320z + 64z 2
Examples

A. If f (x) = 3x – 4, f nd f (4).
f (4) = 3(4) – 4 Replace x with 4. Your Turn!
= 12 – 4 Multiply. Function Values
= 8 Subtract.
A. If f(x) = 2x + 5, f nd f(3).

Answer f (4) = 8 A. 8
B. 7
C. 6
D. 11
B. If f(x) = 3x – 4, f nd f(–5).
f(–5) = 3(–5) – 4 Replace x with –5.
Answer
= –15 – 4 Multiply.
= –19 Subtract.

B. If f (x) = 2x + 5, f nd f (–8).
Answer f(–5) = –19
A. –3
B. –11
C. 21
D. –16
Nonlinear Function Values

Example Answer

Nonlinear Function Values
PHYSICS
2
The function h(t ) = 160t + 16t represents the The function h(t) = 180 – 16t represents the height
2
height of an object ejected downward from an of a ball thrown from a cliff that is 180 feet above
airplane at a rate of 160 feet per second. the ground. Find the value h(3z).

A. Find the value h(3).
2
A. 180 – 16z ft
h(3) = 160(3) + 16(3) 2 Replace t with 3. B. 180 ft
C. 36 ft
= 480 + 144 Multiply.
2
D. 180 – 144z ft
= 624 Simplify.

Answer h(3) = 624
Answer

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