calculus help!?

Question

Find the horizontal and vertical asymptotes of the graph of the function f(x) = [(5 x^2 −2 x + 5)/(x^2 −3)].
The horizontal asymptote is y=?
There are two vertical asymptotes. The one on the left of zero is x= ?
The asymptote on the right of zero is x= ?

Answer 1

f(x) = [(5x^2 - 2x + 5) / (x^2 - 3)]

To find the vertical asymptotes, we must determine what makes the denominator equal to 0.

x^2 - 3 = 0
x^2 = 3. Take the square root of both sides, remembering to include "plus or minus".
x = +/- sqrt(3)

Therefore, x = sqrt(3) and x = -sqrt(3) will be our vertical asymptotes.

To find the horizontal asymptotes, we take the limit as x approaches infinity and the limit as x approaches negative infinity.

lim [ (5x^2 - 2x + 5) / (x^2 - 3) ]
x -> infinity

Divide each term by the highest power of x. In this case, it's x^2. Doing so gives us (after the reduction)

lim [ (5 - 2/x + 5/x^2) / (1 - 3/x^2) ]
x -> infinity

Remember that c/x^n, for n >= 1 and c a constant, as x appraoches infinity is equal to 0. Therefore, evaluating our limit gives us

(5 - 0 + 0) / (1 - 0) = 5/1 = 5

y = 5 is our horizontal asymptote.

If we take the limit as x approaches negative infinity, we will get the same result.

Vertical asymptotes: x = -sqrt(3) and x = sqrt(3).
Horizontal asymptote: y = 5.

Answer 2

The two vertical asymptotes are the square root of 3 and negative square root of 3 since these are the two x values that will cause the denomiator to be zero, driving the y value towards infinitiy.

The horizontal asymptote is y=5. As x grows large, the y value approaches 5.

Answer 3

The horizontal is y=5.
The two vertical asymptotes are x=-sqrt(3) and x=sqrt(3).

Answer 4

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