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.

Short 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$-2$.

Step-by-step solution

Part (a) Step 1. Substitute x=-6 in H(x)=(12)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,

$x=-2$