Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 2.1
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Exercise 72 Page 81

Practice makes perfect
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, 9) & 9=ab^2 [0.4em] [-0.8em] ( 4, 324) & 324=ab^4 [0.4em] If we combine these, we get a system of equations which we can be solved by using the Substitution Method.
9=ab^2 & (I) 324=ab^4 & (II)
â–Ľ
(I): Solve for a
9b^2=a 324=ab^4
a= 9b^2 324=ab^4
a= 9b^2 324= 9b^2* b^4
â–Ľ
(II): Solve for b
a= 9b^2 324= 9b^4b^2
a= 9b^2 324=9b^2
a= 9b^2 36=b^2
a= 9b^2 b^2=36
a= 9b^2 b=± 6

(II): b > 0

a= 9b^2 b=6
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= 9b^2 b=6
a= 9 6^2 b=6
â–Ľ
(I): Evaluate right-hand side
a= 936 b=6
a= 14 b=6
Now we can complete the equation. y= 14(6)^x
b Like in Part A, we have two points through which the exponential function passes. Notice though that one point is on the y-axis, (0,12). If we substitute this point into the general form of the exponential function, we can solve for a since a base raised to the zeroth power equals 1.
12=ab^0 ⇔ a=12When we know that a=12, we can find b by substituting the second point into the function and solving for b.
y=12b^x
40=12b^(- 1)
â–Ľ
Solve for b
40=12* 1/b
40b=12
b=12/40
b=3/10
Now we can complete the equation. y=12(3/10)^x