# How do you find the domain, x intercept and vertical asymptotes of h(x)=log_4(x-3)?

##### 1 Answer
Aug 15, 2017

$\text{Domain} : x > 3$

$x \text{-intercept} : x = 4$

$\text{Vertical asymptote:}$ $x = 3$

#### Explanation:

We have: $h \left(x\right) = {\log}_{4} \left(x - 3\right)$

For the domain of this function, we must consider the argument of the logarithm.

The argument of any logarithmic number must be greater than $0$:

$R i g h t a r r o w x - 3 > 0$

$\therefore x > 3$

The domain of $h \left(x\right)$ is $x > 3$.

Then, let's set $h \left(x\right) = 0$:

$R i g h t a r r o w {\log}_{4} \left(x - 3\right) = 0$

Using the laws of logarithms:

$R i g h t a r r o w x - 3 = {4}^{0}$

$R i g h t a r r o w x - 3 = 1$

$\therefore x = 4$

The $x$-intercept of $h \left(x\right)$ is $x = 4$.

Now, we have already determined that the domain of $h \left(x\right)$ is $x > 3$.

This means that the graph of $h \left(x\right)$ will never touch the line at $x = 3$

The vertical asymptote of $h \left(x\right)$ is $x = 3$.