Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
3. Section 10.3
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Exercise 119 Page 508

Practice makes perfect
a To graph the function, we have to calculate a few ordered pairs through which the function's graph passes. To do so, we choose a few x-values and calculate their corresponding y-values.
x 2(0.75)^x y
- 2 2(0.75)^(- 2) 3.56
- 1 2(0.75)^(- 1) 2.67
0 2(0.75)^0 2
1 2(0.75)^1 1.5
2 2(0.75)^2 1.125
3 2(0.75)^3 0.84

We now plot the points and connect them with a smooth curve.

b The general form of an exponential function has a specific format.
f(x)=ab^xIn this equation, a is the initial value and b is the multiplier. If we divide 52.9 by its preceding value, we can calculate the multiplier. 52.9/23=2.3 The multiplier is 2.3. When we know the multiplier, we can substitute one of the ordered pairs into the function rule and solve for a. Let's use the pair (1,23).
f(x)=a(2.3)^x
f( 1)=a(2.3)^1
23=a(2.3)^1
â–Ľ
Solve for a
23=a(2.3)
10=a
a=10
The function is f(x)=10(2.3)^x.
c From the equation, we see that the initial value is 75 and the multiplier is 0.85.

f(x)= 75( 0.85)^x A multiplier of 0.85 corresponds to a decrease of 15 %. A possible context based on this equation could be that you have $75 deposited in your bank account and spend 15 % of what is left in your bank account each month.