Chapter 5.3, Problem 90AYU

To determine

To find: An exponential function with horizontal asymptote

y = -3 whose graph contains the points (0,-2) and (-2,1).

Answer to Problem 90AYU

Solution:

  $f\left(x\right)={\left(\frac{1}{2}\right)}^x-3$

Explanation of Solution

Given:

Asymptote of the graph is $y=-3$ and the points are $\left(0,-2\right)$ and $\left(-2, 1\right)$ .

Calculation:

  $y=a{b}^x+c$ , where $c$ is asymptote

Therefore $y=a{b}^x-3$ ----------(1)

Substitute $\left(0, -2\right)$ in (1)

  $-2=a{b}^0-3$

  $-2=a-3$

  $a=1$ ----------------(2)

Substitute $\left(-2, 1\right)$ in (1)

  $1=a{b}^{-2}-3$

  $1=\left(1\right)b^{-2}-3$ by(2)

  $1+3=b^{-2}$

  $b^{-2}=4$

  $\frac{1}{b^2}=4$

  $b^2=\frac{1}{4}$

  $b=\frac{1}{2}$ -----------------(3)

Substitute (2) and (3) in (1)

  $y={\left(\frac{1}{2}\right)}^x-3$