Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 12.2
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Exercise 109 Page 686

Practice makes perfect
a 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 a function.
y=ab^x Since we want the points to lie on the graph, they must both satisfy this equation. By substituting these points into the formula we get two equations.
Point y = ab^x
( 1, 6) 6=ab^1
( 1, 6) 6=ab^1
( 5, 30.375) 30.375=ab^5
If we combine these equations, we get a system of equations which we can solve for a and b.
6=ab^1 & (I) 30.375=ab^5 & (II)
â–Ľ
(I): Solve for a
6=ab 30.375=ab^5
ab=6 30.375=ab^5
a= 6b 30.375=ab^5
a= 6b 30.375=( 6b)b^5
â–Ľ
(II): Solve for b
a= 6b 30.375= 6b^5b
a= 6b 30.375=6b^4
a= 6b 6b^4=30.375
a= 6b b^4=5.0625
a= 6b b=±1.5

(II): b > 0

a= 6b b=1.5
Having solved for b, we can substitute this into the first equation to find a.
a= 6b b=1.5
a= 6 1.5 b=1.5
a=4 b=1.5
Finally, we can write the full equation of the exponential function. y=4(1.5)^x
b The b-value from Part A tells us how fast the function increases — since b = 1.5, increasing the value of x by one makes the value of the function 1.5 times greater. Let's take a look at a following example with x = 0.

As we can see, each time the value of x was increased by 1, the value of function increased 1.5 = b times.

c Let's think of a situation that when each time a certain interval passes, the amount of something gets multiplied by a given value. One such example would be the amount of debt we owe if we borrowed $4 with 0.5 annual interest.