Exercise 63 Page 312

We want to write an exponential function so that the slope from the point (0, f(0)) to the point (2, f(2)) is equal to 12. Let's move the points so that the slope between the points is 12 and that the x-coordinates of the points are 0 and 12. Now, we have an idea about how the graph changes as the points change. Using the Slope Formula, we can write the following equation. m=y_2-y_1/x_2-x_1 ⇒ 12=f(2)-f(0)/2-0Let's simplify it.

12=f(2)-f(0)/2-0

IdPropAdd

Identity Property of Addition

12=f(2)-f(0)/2

MultEqn

LHS * 2=RHS* 2

24=f(2)-f(0)

Since f(x) is an exponential function, it has the form f(x)=a(b)^x. Then, f(2)=a(b)^2 and f(0)=a(b)^0. f(x)=a(b)^x ⇒ lf(0)= a(b)^0 f(2)= a(b)^2 Let's substitute the equivalent expressions.

24=f(2)-f(0)

SubstituteExpressions

Substitute expressions

24= a(b)^2- a(b)^0

▼

Factor RHS

ExponentZero

a^0=1

24=a(b)^2-a(1)

IdPropMult

Identity Property of Multiplication

24=a(b)^2-a

FactorOut

Factor out a

24=a(b^2-1)

FacDiffSquares

a^2-b^2=(a+b)(a-b)

24=a(b+1)(b-1)

If we let a be equal to, for example, 1, then we have the following equation. 24 =a(b+1)(b-1) ⇓ 24 =(b+1)(b-1) We need to find two factors whose difference is 2 and product is 24. For example, 6 and 4. 24 = (b+1)_6 (b-1)_4 ⇒ b=5 Therefore, b is 5. We can now substitute a=1 and b=5 into the function form. f(x) = a (b)^2 ⇒ f(x) = 1(5)^x This exponential function satisfies the desired condition. Please note that we can find different functions for this exercise.

Checking Our Answer

Checking Our Answer

Let's first find the points on the exponential function when x=0 and x=2.

	f(x)= 5^x
Operation	x=0	x=2
Substitution	f( 0)= 5^0	f( 2)= 5^2
Point	(0,1)	(2,25)

Now, we can calculate the slope from (0,1) to (2,25).

m = y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( 0,1) & ( 2,25)

m = 25- 1/2- 0

SubTerm

Subtract term

m = 24/2

CalcQuot

Calculate quotient

m = 12

The slope is equal to 12, so we found a correct function.