Exponential functions

Question

Determine the equation of the exponential function whose graph contains points (-1,3) and (2,8/9)

Answer 1

 

We want an equation of the form

y₁ = P₀a^x₁

y₂ = P₀a^x₂

 

y₂/y₁ = P₀a^x₂/P₀a^x₁ = a^x₂/a^x₁ = a^x₂-x₁

 

Now raise both sides of the equation to the power 1/(x₂-x₁)

 

(y₂/y₁)^{1/(x₂-x₁)} = a^{[(x₂-x₁)/(x₂-x₁)]} = a

 

Let's solve for a by filling in the (x,y) values given.

 

[(8/9)/3]^1/[2-(-1)]

(8/27)^1/3

2/3

a = 2/3

 

so,

y = P₀(2/3)^x

 

Let's use one of the (x,y) points given to solve for P₀.

3 = P₀(2/3)^-1 P₀/(2/3) = 3P₀/2

P₀ = 2

 

Finally,

y = 2 (2/3)^x

 

Answer 2

An exponential function has the form y = A(B)^x.

 

Since (-1,3) is on the graph, 3 = A(B^-1).  So, A = 3B

 

Since (2,8/9) is on the graph, 8/9 = A(B)²

 

So, 8/9 = 3B(B²)

 

     8/9 = 3B³

 

       B³ = 8/27

 

       B = 2/3

 

       A = 3B = 2

 

Equation is y = 2(2/3)^x