# How do you graph f(x)=7/(x+4) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jan 15, 2018

Holes : None

Vertical asymptote : $x = - 4$

Horizontal asymptote : $y = 0$

$x$-intercept : None

$y$-intercept : $\left(0 , \frac{7}{4}\right)$

#### Explanation:

Holes are literally what they mean—a hole in a graph. It's when both numerator and denominator of the equation have an exact same factor. For example, in the equation $f \left(x\right) = \frac{3 \left(x + 1\right)}{x + 1}$, there would be a hole at $- 1$.

In our equation $f \left(x\right) = \frac{7}{x + 4}$, there are no factors on both numerator and denominator of the function, so there are no holes in the graph.

Vertical asymptotes are vertical lines where the graph approaches and get closer to but NEVER touches.

To find whether there is/are vertical asymptote(s), we set the denominator of the function equal to $0$, like this:
Our denominator of the function is $x + 4$, so we do this:
$x + 4 = 0$
$x = - 4$

As said earlier, vertical asymptotes are vertical lines, meaning they start with $x =$.

Horizontal asymptotes are horizontal lines where the graph approaches; however, it can be crossed.

The following are the rules for finding horizontal asymptotes:
In the following,
Let m be the degree of the numerator.
Let n be the degree of the denominator.

1. if m > n , then there is no horizontal asymptote

2. if m = n , then the horizontal asymptote is dividing the coefficients of the numerator and denominator

3. if m < n , then the horizontal asymptote is y=0

From the equation $f \left(x\right) = \frac{7}{x + 4}$, we can see that the degree of the denominator is higher than the degree of the numerator (there's an $x$ on the bottom but no $x$ on top). This means that the horizontal asymptote is $y = 0$.

X-intercepts are where the graph touches the x-axis. It's sort of similar to finding the vertical asymptote, but we look at the numerator instead.
We set the numerator equal to $0$:
$7 = 0$ BUT THAT IS NEVER TRUE! $7 \ne 0$
This means that there are no x-intercepts.

Y-intercepts are the values of $y$ when you plug in $0$ for $x$ or where the graph touches the y-axis.
As said, let's plug in $0$ for the $x$ values:
$f \left(0\right) = \frac{7}{x + 4}$
$f \left(0\right) = \frac{7}{0 + 4}$
$f \left(0\right) = \frac{7}{4}$

So there is an y-intercept at $\left(0 , \frac{7}{4}\right)$.

If you need more help or want to watch a video, feel free to watch this:

Sorry, I know it's kind of long!
Hope this helps!