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

1 Answer
Dec 29, 2016

Answer:

Horizontal : #larr y = -8 rarr#. Vertical + #uarr x = - 5 darr#
The graph is a rectangular hyperbola. x-intercept ( y = 0 ): #-11/2#. y-intercept ( x = 0 ). #-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)(x)+(R_l(x))/(S_n(x))#,

where #n>=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 = -8-4/(x+5)#

n = m = 1.

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

So, the asymptotes are

#y = -8 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

#(y-mx+a)(y+x/m+b)=0#