Question

R(x)=x^2 + 14x + 24/ x^2 + 13x + 12 find the horizontal or oblique asymptotes

Answer 1

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Answer 2

y = 1 right?

Answer 3

no oblique asymptotes since the degree of the numerator is the same as the degree of the denominator. horizontal is y = 1, the ratio of the leading coefficients

Answer 4

vertical asymptotes set denominator = 0 and solve

Answer 5

\[x^2+13x+12=0\]
\[(x+12)(x+1)=0\]
\[x=-12\]
\[x=1\] are your vertical asymptotes

Answer 6

thinking of course of
\[x=-12\] as the vertical line , not the number

Answer 7

and of course i made a typo. should be \[x=-1\] for the other one

Answer 8

amistre, i got the bike!

Answer 9

how do you show the steps for the above problem

Answer 10

do those zeros have any tops that cancel?

Answer 11

yay!! ... i got that pitman arm that needs replaced....

Answer 12

good question. if so i am dead wrong

\[\frac{x^2+14x+24}{x^2+13x+12}\]
\[=\frac{(x+12)(x+2)}{(x+12)(x+1)}=\frac{x+2}{x+1}\]

Answer 13

so i was completely wrong. horizontal asymptote still y = 1
vertical only at x = -1 and a hole at x = -12 sorry

Answer 14

thanks amistre for setting me straight!

Answer 15

question is what kind of moron would write
\[\frac{x^2+14x+24}{x^2+13x+12}\] when the just meant

\[\frac{x+2}{x+1}\]?

Answer 16

lol .... sometimes that data follows the rational expression f(x)/g(x); but the graph of it can be simplified by (x+2)/(x+1)

Answer 17

of course with the hole at -12