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

Exercise 175 Page 116

a The general form of an exponential function is written in the following format.
y=ab^x To find the equation, we need to determine a and b. From the exercise, we know that the function passes through two points. This means we can substitute both of these points in the function creating two equations. |c|c| [-0.8em] Point & y=ab^x [0.4em] [-0.8em] ( 2, 99) & 99=ab^2 [0.4em] [-0.8em] ( 6, 8019) & 8019=ab^6 [0.4em] If we combine these, we get a system of equations which we can be solved by using the Substitution Method.
99=ab^2 & (I) 8019=ab^6 & (II)
(I): Solve for a
99b^2=a 8019=ab^6
a= 99b^2 8019=ab^6
a= 99b^2 8019= 99b^2(b^6)
(II): Solve for b
a= 99b^2 8019= 99b^6b^2
a= 99b^2 8019=99b^4
a= 99b^2 81=b^4
a= 99b^2 b^4=81
a= 99b^2 b=± 3

(II): b > 0

a= 99b^2 b=3
Notice that b must be non-negative since we cannot have a negative base in an exponential function. To find a, we substitute the value of b back into the first equation and evaluate.
a= 99b^2 b=3
a= 99 3^2 b=3
(I): Evaluate right-hand side
a= 999 b=3
a=11 b=3
Now we can complete the equation. y=11(3)^x
b Like in Part A, we have to substitute the known points into the general form of an exponential function and then solve for a and b.
|c|c| [-0.8em] Point & y=ab^x [0.4em] [-0.8em] ( -1, 50) & 50=ab^(-1) [0.4em] [-0.8em] ( 2, 25.6) & 25.6=ab^2 [0.4em] If we combine these, we get a system of equations which can be solved by using the Substitution Method.
50=ab^(- 1) & (I) 25.6=ab^2 & (II)
(I): Solve for a
50=a( 1b) 25.6=ab^2
a( 1b)=50 25.6=ab^2
a=50b 25.6=ab^2
a=50b 25.6=( 50b)b^2
(II): Solve for b
a=50b 25.6=50b^3
a=50b 0.512=b^3
a=50b b^3=0.512
a=50b b=0.8
To find a, we substitute the value of b back into the first equation and evaluate.
a=50b b=0.8
a=50( 0.8) b=0.8
a=40 b=0.8
Now we can complete the equation. y=40(0.8)^x