# How do you graph f(x)=4(3/4)^(x+1)-5 and state the domain and range?

Nov 8, 2017

See below.

#### Explanation:

The y axis intercept occurs where $x = 0$, so:

$4 {\left(\frac{3}{4}\right)}^{0 + 1} - 5 = - 2$ coordinate $\left(0 , - 2\right)$

There are no restrictions on $x$ so:

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

${\lim}_{x \to \infty} 4 {\left(\frac{3}{4}\right)}^{x + 1} - 5 = - 5$ ( y = -5 is a horizontal asymptote )

For $x < - 1$

$4 {\left(\frac{3}{4}\right)}^{x + 1} - 5$ becomes:

$\frac{4}{{\left(\frac{3}{4}\right)}^{x + 1}} - 5$

as $x \to \infty$ ,${\left(\frac{3}{4}\right)}^{x + 1} \to 0$

So:

${\lim}_{x \to - \infty} 4 {\left(\frac{3}{4}\right)}^{x + 1} - 5 = \infty$

Range is:

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

Graph: