Exercise 112 Page 358

We want to write an exponential function with the given y-intercept and multiplier. Let's consider the general form for this type of function.

y=ab^xIn the formula above, the base b can be interpreted as the constant multiplier. Since we know that the multiplier equals 0.8, we can substitute b=0.8. y=ab^x ⇒ y=a(0.8)^x We know that (0,2) is the y-intercept. We want this point to lie on the graph, so it must satisfy the above equation. Let's substitute (0,2) into this formula.

y=a(0.8)^x

SubstituteII

x= 0, y= 2

2=a(0.8)^0

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Solve for a

ExponentZero

a^0=1

2=a(1)

IdPropMult

Identity Property of Multiplication

2=a

RearrangeEqn

Rearrange equation

a=2

Now we can write the full equation of the exponential function. y= a(0.8)^x ⇒ y= 2(0.8)^x

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 (0,3.5) into the above formula.

y=ab^x

SubstituteII

x= 0, y= 3.5

3.5=ab^0

▼

Solve for a

ExponentZero

a^0=1

3.5=a(1)

IdPropMult

Identity Property of Multiplication

3.5=a

RearrangeEqn

Rearrange equation

a=3.5

Now we can partially write our equation. y= ab^x ⇒ y= 3.5b^x Next, we will substitute the second given point, (2,31.5), into the above equation and solve for b.

y=3.5b^x

SubstituteII

x= 2, y= 31.5

31.5=3.5b^2

▼

Solve for b

DivEqn

.LHS /3.5.=.RHS /3.5.

31.5/3.5=b^2

CalcQuot

Calculate quotient

9=b^2