Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
Chapter Closure

Exercise 175 Page 366

The general equation of an exponential function is y=ab^x. The asymptote at y=20 suggests that the function is translated.

y=8(1/4)^x+20

We want to write an exponential function for the graph that passes through the given points. Let's consider the general form for this type of function. y=ab^x Such functions have horizontal asymptote at y=0. However, we are told that the function has a horizontal asymptote at y=20. This means that the given function is translated 20 units up. y=ab^x+20Now, since we want the points to lie on the graph they must satisfy this equation. Let's substitute (1,22) into the above formula.
y=ab^x+20
22=ab^1+20
Solve for a
22=ab+20
2=ab
ab=2
Next, we will substitute the second given point, (3,20.125).
y=ab^x+20
20.125=ab^3+20
Solve for b
0.125=ab^3
ab^3=0.125
We received two equations, so to find the values of a and b we need to solve the system of equations. ab=2 & (I) ab^3=0.125 & (II) The first equation says that the product of a and b is 2. Notice that the second equation contain the same expression multiplied by b^2. ab=2 ab^3=0.125 ⇒ ab=2 ab* b^2=0.125 This allows us to substitute the value of ab to the second equation and solve for b. Keep in mind that b is the base of the exponential function, so it has to be positive.
ab=2 & (I) ab* b^2=0.125 & (II)
ab=2 2b^2=0.125
(II):Solve for b
ab=2 2b^2=1/8
ab=2 b^2=1/8÷ 2
ab=2 b^2=1/16
ab=2 sqrt(b^2)=sqrt(1/16)
ab=2 b=sqrt(1/16)
ab=2 b=1/4
Once we know the value of b, let's substitute it to the first equation to find a.
ab=2 b=1/4
a* 14=2 b= 14
(I): Solve for a
a/4=2 b=1/4
a=8 b=1/4
Finally, we can write the full equation of the exponential function. y= a b^x+20 ⇒ y= 8( 1/4)^x+20