# How do you graph y=-4/(x+5)-8 using asymptotes, intercepts, end behavior?

Dec 29, 2016

Horizontal : $\leftarrow y = - 8 \rightarrow$. Vertical + $\uparrow x = - 5 \downarrow$
The graph is a rectangular hyperbola. x-intercept ( y = 0 ): $- \frac{11}{2}$. y-intercept ( x = 0 ). $- \frac{44}{5}$.

#### Explanation:

graph{(y+8)(x+5)+4=0 [-80, 80, -40, 40]} General method for the benefit of students:

If the equation can be presented in the quotient-remainder form

$y = {Q}_{n - m} \left(x\right) + \frac{{R}_{l} \left(x\right)}{{S}_{n} \left(x\right)}$,

where $n \ge m$ and l is utmost n-m, and Q, R and S are polynomials

of degrees indicated by the suffixes,

then y = Q and S = 0, give the equations to the asymptotes.

y = Q is a slant straight line, if n=m+1,

It is a horizontal straight line, if m = n.

If S = 0 gives horizontal asymptotes y - real zero of S.

In other cases, we get asymptotic curves.

Here,

$y = - 8 - \frac{4}{x + 5}$

n = m = 1.

Q =-8 and S = x+5..

So, the asymptotes are

$y = - 8 \mathmr{and} x + 5 = 0$.

Short method for this problem:

The equation can be remodelled to the form

(y+8)(x+5)=-4 that represents a rectangular hyperbola ( RH ), with

asymptotes (y+8)(x+5)=0, by comparison with the standard form

((y-mx-a)(y+x/m+b)=c that represents a RH, with asymptotes

$\left(y - m x + a\right) \left(y + \frac{x}{m} + b\right) = 0$