Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 7.2
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Exercise 87 Page 344

Practice makes perfect
a

We want to write an exponential function for the graph that has an initial value of 2 and passes through the given point. Let's consider the general form for this type of function.

y=ab^xSince 2 is the initial value, the graph of the given function starts at the point (0,2). We want the points to lie on the graph, so they must satisfy the above equation. Let's substitute (0,2) into this formula.

y=ab^x
2=ab^0
â–¼
Solve for a
2=a(1)
2=a
a=2

Now we can partially write our equation. y= ab^x ⇒ y= 2b^x Next, we will substitute the second given point, (3,128), into the above equation and solve for b.

y=2b^x
128=2b^3
â–¼
Solve for b
64=b^3
sqrt(64) = b
4 = b
b=4

Finally, we can write the full equation of the exponential function. y=2b^x ⇒ y=2(4)^x

b

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^xSince we want the points to lie on the graph, they must satisfy this equation. Let's substitute (0,4) into the above formula.

y=ab^x
4=ab^0
â–¼
Solve for a
4=a(1)
4=a
a=4

Now we can partially write our equation. y= ab^x ⇒ y= 4b^x Next, we will substitute the second given point, (2,1), into the above equation and solve for b.

y=4b^x
1=4b^2
â–¼
Solve for b
1/4=b^2
sqrt(1/4) = b
sqrt(1)/sqrt(4) = b
1/2 = b
b=1/2

Finally, we can write the full equation of the exponential function. y=4b^x ⇒ y=4 (1/2)^x