Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 9.2
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Exercise 107 Page 570

Practice makes perfect
a We want to draw a graph of the given exponential function.
f(x)=2( 0.75)^x

Because the base of the function is greater than 0 but less than 1, we know that this is an exponential decay function. To draw the graph, we will start by making a table of values.

x 2(0.75)^x y=2(0.75)^x
- 3 2(0.75)^(- 3) 4.740740...
- 2 2(0.75)^(- 2) 3.555555...
- 1 2(0.75)^(- 1) 2.666666...
0 2(0.75)^0 2
1 2(0.75)^1 1.5
2 2(0.75)^2 1.125

The ordered pairs ( - 3, 4.741), ( - 2, 3.555), ( - 1, 2.666), ( 0, 2), ( 1, 1.5), and ( 2, 1.125) all lie on the graph of the function. Now we can plot and connect these points with a smooth curve.

b We want to write an exponential function for the graph that passes through the points listed in the given table.
x f(x)
1 23
2 52.9
3 121.67
4
Let's consider the general form for this type of function. y=ab^xSince we want the points to lie on the graph, they must satisfy this equation. Let's substitute (1,23) into the formula.
y=ab^x
23=ab^1
23=ab
ab = 23
This gives an equation relating a with b, but we do not know either of these values yet. Let's substitute 2,52.9) into the function to get another equation for the value of ab.
y=ab^x
52.9=ab^2
ab^2 = 52.9
Now we have created a system of equations for a and b. ab=23 & (I) ab^2=52.9 & (II) Notice that if we divide Equation (II) by Equation (I), we can solve for the b-value. Let's do it!
ab^2/ab = 52.9/23
ab(b)/ab = 52.9/23
b = 52.9/23
b = 2.3
Now we can partially write our equation. y=a b^x ⇒ y=a( 2.3)^x Next, we will substitute the 2.3 for b in Equation (I) and solve for a.
ab=23
a(2.3)=23
a = 10
Finally, we can write the full equation of the exponential function. y= a(2.3)^x ⇒ y= 10(2.3)^x
c First, let's recall the formula for the exponential decay using the percent decrease rate.
y = a(1- p)^xHere, a is the initial amount and p is the percent decrease. We are told that the current population of Flood River City is 42 000 people and is expected to decrease by 25 % — or 0.25 — each year for the next five years. Let's write our function using this information. y = 42 000(1 - 0.25)^x Since we want to find the population after five years, we will substitute 5 for x in our equation and solve for y. Let's do it!
y = 42 000(1- 0.25)^x
y = 42 000(1- 0.25)^5
y = 42 000(0.75)^5
y = 42 000( 0.237304...)
y= 9966.796875...
y ≈ 9967
Therefore, after five years, the population of Flood River City should be about 9967 people.
d We want to identify the annual multiplier b and the percent increase in the given situation. To do so, let's first recall the general form of an exponential growth function.
y= a * b^x, where a>0 and b > 1 Here, a is the initial amount and the base b is the multiplier, which equals 1 plus the decimal form of the percent increase. We want to find these values based on the given information. Here, we can say that the value of share of Orange stock in 2000, $25, is the initial amount. a = $25This allows us to write a partial version our equation. y = 25 * b^x Now let's turn our attention back to the information provided. We know that in 2010, 10 years after 2000, the shares were worth $60. Let's write this as an equation for b, the annual multiplier. 60 = 25 * b^(10) Let's solve the equation for b.
60 = 25 * b^(10)
Solve for b
2.4 = b^(10)
sqrt(2.4) = b
1.091493... = b
1.09 ≈ b
b ≈ 1.09
We only took the principal root because the multiplier must be a non-negative number. Therefore, the annual multiplier is about 1.09. Now, recall that the multiplier equals 1 plus the percent increase expressed as a decimal. b = 1 + p Let's solve the equation above for the percent increase p.
b = 1 + p
1.09 = 1 + p
0.09 = p
p = 0.09
Therefore, the percent increase is about 0.09, or 9 %.