Question

Given the following rational function:

(a) state the domain.

(b) find the vertical and horizontal asymptotes, if any.

(c) find the oblique asymptotes, if any.

(d) submit a graph of this function to the Dropbox.

Please show all of your work.
f(x)=(x^2+ 2x-8)/(x^2 + 7x + 12)

Answer 1

Domain:
all values of x except -4 and -3

Answer 2

can you show me the steps on how you got that ?

Answer 3

f(x)=(x^2+ 2x-8)/(x^2 + 7x + 12)

factors to [(x-2 )(x+4)] / {(x + 3) ( x + 4)] = f(x)

Answer 4

The horizontal asymptote has equation y = 1.

As x increases without bound, y = (x-2)/(x+3) approaches 1.

Answer 5

There may be just one vertical asymptote which would be x = -3.

[(x-2 )(x+4)] / {(x + 3) ( x + 4)] = f(x)

The (x+4) terms divide out giving (x - 2) / (x + 3) = f(x).

The "problem" at x = -4 may be a removeable discontinuity.

See graph at: http://www.wolframalpha.com/input/?i=y+%3D%28x%5E2%2B+2x-8%29%2F%28x%5E2+%2B+7x+%2B+12%29 while I read check on removeable discontinuties.

Answer 6

Removeable Discontinuity at (-4,6) --> shows as "hole" on graph
y-intercept (0,-2/3)
x-intercept (2,0)
vertical asymptote x = -3
horizontal asymptote y = 1
no oblique asymptote




Removeable Discontinuity at (-4,6)