# How do you graph, find any intercepts, domain and range of f(x)=(1/4)^(x+4)-3?

Dec 3, 2017

See below.

#### Explanation:

y axis intercept occurs when $x = 0$

$y = {\left(\frac{1}{4}\right)}^{0 + 4} - 3 = \frac{1}{256} - 3 = \textcolor{b l u e}{- \frac{767}{256}}$

x axis intercept occurs when $y = 0$

${\left(\frac{1}{4}\right)}^{x + 4} - 3 = 0$

${\left(\frac{1}{4}\right)}^{x + 4} = 3$

$\left(x + 4\right) \ln \left(\frac{1}{4}\right) = \ln 3 \implies x = \frac{\ln 3}{\ln \left(\frac{1}{4}\right)} - 4 \approx \textcolor{b l u e}{- 4.792}$

There are no constraints on $x$ so domain is:

$\textcolor{b l u e}{\left\{x \in \mathbb{R}\right\}}$

as $x \to \infty$ , $\textcolor{w h i t e}{888} {\left(\frac{1}{4}\right)}^{x + 4} \to 0$

So:

as $x \to \infty$ , $\textcolor{w h i t e}{888} {\left(\frac{1}{4}\right)}^{x + 4} - 3 \to - 3$

The line $\textcolor{b l u e}{y = - 3} \textcolor{w h i t e}{88}$ is a horizontal asymptote.

For $x < - 4$

${\left(\frac{1}{4}\right)}^{x + 4}$ becomes $\textcolor{w h i t e}{888} \frac{1}{{\left(\frac{1}{4}\right)}^{x + 4}}$

So:

as $x \to - \infty$ , $\textcolor{w h i t e}{888} \frac{1}{{\left(\frac{1}{4}\right)}^{x + 4}} \to \infty$

and:

$x \to - \infty$ , $\textcolor{w h i t e}{888} \frac{1}{{\left(\frac{1}{4}\right)}^{x + 4}} - 3 \to \infty$

$\textcolor{b l u e}{\left\{y \in \mathbb{R} : - 3 < y < \infty\right\}}$

Graph: