# How do you graph y=(x^2-5x-36)/(3x) using asymptotes, intercepts, end behavior?

Dec 1, 2016

See the explanation below

#### Explanation:

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{0\right\}$

As $x \ne 0$

So, $x = 0$ is a vertical asymptote.

The numerator is ${x}^{2} - 5 x - 36$

$= \left(x + 4\right) \left(x - 9\right)$

So, the intercepts when $y = 0$

are $\left(- 4 , 0\right)$ and $\left(9 , 0\right)$

For the limits $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} {x}^{2} / \left(3 x\right) = {\lim}_{x \to \pm \infty} \frac{x}{3} = \pm \infty$

${\lim}_{x \to {0}^{-}} y = \frac{{0}^{+} + 5 - 36}{3 \cdot {0}^{-}} = + \infty$

${\lim}_{x \to {0}^{+}} y = \frac{{0}^{+} - 5 - 36}{3 \cdot {0}^{+}} = - \infty$

Now you can draw your graph

graph{(x^2-5x-36)/(3x) [-32.47, 32.47, -16.24, 16.25]}

Dec 1, 2016

Asymptotes: $x = 0 \mathmr{and} x - 3 y - \frac{5}{3} = 0.$ x-intercepts by the two branches: -4 and 9. As $x \to 0 , y \to \pm \infty$ and the other asymptote tends to reach the hyperbola, at infinite distance in ${Q}_{1} \mathmr{and} {Q}_{3}$

#### Explanation:

graph{x^2-3xy-5x-36=0 [-20, 20, -10, 10]} The second degree equation $a {x}^{2} + 2 h x y + b {y}^{2} + \ldots = 0$

represents a hyperbola, when $a b - {h}^{2} < 0$ and the second degree

terms reveal the first degree terms in the equations of its

asymptotes.

Here, after cross multiplication,

$x \left(x - 3 y\right) - 5 x - 36 = 0$. Reconstructing as

((x+a)(x-3y+b)+c=0, we find that

$a = 0 , b = - \frac{5}{3} \mathmr{and} c = - 36$.

So, the asymptotes are given by

x=0 and x-3y-5/3=0, meeting at the center C(0, -5/9)#.

x-intercepts by the two branches: -4 and 9.

As $x \to 0 , y \to \pm \infty$ and the other asymptote tends to reach the

hyperbola, at infinite distance in ${Q}_{1} \mathmr{and} {Q}_{3}$