# How do you graph, find the intercepts and state the domain and range of f(x)=1/2(2^x-8)?

Jul 29, 2018

The x-intercept is $\left(3 , 0\right)$. The y-intercept is $\left(0 , - \frac{7}{2}\right)$.
The domain is $x \in \mathbb{R}$. The range is $y \in \left(- 4 , + \infty\right)$

#### Explanation:

The function is

$y = \frac{1}{2} \left({2}^{x} - 8\right)$

The intercepts are

when $x = 0$

$\implies$, $y = \frac{1}{2} \left({2}^{0} - 8\right) = \frac{1}{2} \left(1 - 8\right) = - \frac{7}{2}$

The y-intercept is $\left(0 , - \frac{7}{2}\right)$

when $y = 0$

$\implies$, $0 = \frac{1}{2} \left({2}^{x} - 8\right)$

$\implies$, ${2}^{x} - 8 = 0$

$\implies$, ${2}^{x} = {2}^{3}$

$\implies$, $x = 3$

The x-intercept is $\left(3 , 0\right)$

The domain is $x \in \mathbb{R}$

To find the range, proceed as follows

$y = \frac{1}{2} \left({2}^{x} - 8\right)$

$2 y = {2}^{x} - 8$

${2}^{x} = 2 y + 8$

Taking logs

$\ln \left({2}^{x}\right) = \ln \left(2 y + 8\right)$

$x \ln 2 = \ln \left(2 y + 8\right)$

Therefore,

$2 y + 8 > 0$

$2 y \succ 8$

$y \succ 4$

The range is $y \in \left(- 4 , + \infty\right)$

See the graph belox

graph{(y-1/2(2^x-8))(y+4)=0 [-14.24, 14.24, -7.12, 7.12]}