# How do you graph, find any intercepts, domain and range of f(x)=-(3/4)^(x+2)+5?

Apr 8, 2018

See below.

#### Explanation:

$f \left(x\right) = - {\left(\frac{3}{4}\right)}^{x + 2} + 5$

$y$ axis intercepts occur where $x = 0$

$y = - {\left(\frac{3}{4}\right)}^{0 + 2} + 5 = - {\left(\frac{3}{4}\right)}^{2} + 5 = - \frac{9}{16} + 5 = \textcolor{b l u e}{\frac{71}{16}}$

$x$ axis intercepts occur where $y = 0$

$- {\left(\frac{3}{4}\right)}^{x + 2} + 5 = 0$

${\left(\frac{3}{4}\right)}^{x + 2} = 5$

Taking logarithms of both sides:

$\left(x + 2\right) \ln \left(\frac{3}{4}\right) = \ln \left(5\right)$

$x = \textcolor{b l u e}{\frac{\ln \left(5\right)}{\ln \left(\frac{3}{4}\right)} - 2}$

There are no restriction on $x$, so the domain is:

$\left\{x \in \mathbb{R}\right\}$

To find the range we need to see what happens as $x$ approaches $\pm \infty$

as $x \to \infty$, $\setminus \setminus \setminus \setminus \setminus - {\left(\frac{3}{4}\right)}^{x + 2} + 5 \to 5$

as $x \to - \infty$, $\setminus \setminus \setminus \setminus \setminus - {\left(\frac{3}{4}\right)}^{x + 2} + 5 \to - \infty$

So the range is:

$\left\{y \in \mathbb{R} : 5 < y < \infty\right\}$

The graph confirms these findings: