# How do you graph f(x)=(x^3-2x^2-3x)/(4x^2+8x) using holes, vertical and horizontal asymptotes, x and y intercepts?

Mar 20, 2018

Factor then analyze!

#### Explanation:

We can factor both the top and the bottom to get asymptotes, some intercepts, and holes.

${x}^{3} - 2 {x}^{2} - 3 x = x \left({x}^{2} - 2 x - 3\right) = x \left(x - 3\right) \left(x + 1\right)$
$4 {x}^{2} + 8 x = 4 x \left(x + 2\right)$

We see that they both have a zero in common (x=0), which means that $x = 0$ is the one hole. If we consider the equation without the x term, we see that the function WOULD go through $\left(0 , - \frac{3}{8}\right)$, which is where the hole (and technically the y-intercept) is.

Since the bottom also goes to zero at $x = - 2$, that's a vertical asymptote. Since the top goes to zero at $x = 3 , - 1$ those are zeros.

Last thing we need to observe: the top and bottom are different orders. Since the top is a higher order, in the limit, the equation will look like $y = \frac{x}{4}$.

Now we have everything except exact signs.
At $x \rightarrow - \infty$, the value is clearly negative so
$y < 0 \forall x < - 2$.

It then goes toward negative infinity and jumps to positive infinity after -2, until the next zero, i.e.
$y > 0 \forall - 2 < x < - 1$.

At x=-1, it hits zero and switches sign, i.e.
$y < 0 \forall - 1 < x < 0 \mathmr{and} 0 < x < 3$.

After x=3, the sign doesn't switch and the function quickly approaches that line, i.e.
$y > 0 \forall x > 3$ and $y \rightarrow \frac{x}{4}$.

From all of that analysis, you should be able to sketch a plot similar to this: graph{x(x-3)(x+1)/(4x(x+2)) [-13.86, 13.86, -6.93, 6.93]}