Q. 95 - Precalculus Enhanced with Graphing Utilities

Question

Suppose that $H\left(x\right)={\left(\frac{1}{2}\right)}^x-4$

(a) What is $H(-6)$? What point is on the graph of H?

(b) If $H\left(x\right)=12$, what is x? What point is on the graph of H?

(c) Find the zero of H.

Step-by-Step Solution

Answer

Part (a) $H(-6)=60$ and the point on the graph of H is $(-6,60)$.

Part (b) $x=-4$ and the point on the graph of H is $(-4,12)$.

Part (c) The zero of H is $-2$.

Part (a) Step 1. Substitute x = - 6 in H ( x ) = ( 1 2 ) x - 4 .

This gives

$$\begin{gathered} H(-6)={\left(\frac{1}{2}\right)}^{-6}-4 \\ =2^6-4 \\ =64-4 \\ =60 \end{gathered}$$

The point on the graph of the function H is $(-6,60)$.

Part (b) Step 1. Finding x using H ( x ) = 12 .

We have

$$\begin{gathered} {\left(\frac{1}{2}\right)}^x-4=12 \\ {\left(\frac{1}{2}\right)}^x=16 \\ {2}^{-x}=2^4 \end{gathered}$$

On comparing

$x=-4$

The point on the graph of the function H is $(-4,12)$.

Part (c) Step 1. Substitute H ( x ) = 0 .

This gives

$$\begin{gathered} {\left(\frac{1}{2}\right)}^x-4=0 \\ {\left(\frac{1}{2}\right)}^x=4 \\ {2}^{-x}=2^2 \end{gathered}$$

On comparing,

$x=-2$