which is true for the graph of y=(x^2-4)/(x^2+9)?

Question

A. vertical assymptotes at x= +/-3
B. horizontal assymptotes at y= +/-2
C. vertical assymptotes at x= +/-2
D. horizontal assymptotes at y=1

Answer 1

y = (x^2 - 4)/(x^2 + 9)

Vertical asymptotes are found by equating the denominator to 0. However, x^2 + 9 = 0 has no real solutions; as a result, there are no vertical asymptotes.

To obtain the horizontal asymptotes, you need to calculate the limit of the function as x approaches infinity and x approaches negative infinity.

lim ( x^2 - 4 ) / (x^2 + 9)
x -> infinity

Divide everything by x^2,

lim (1 - 4/x^2) / (1 + 9/x^2)
x -> infinity

And now we evaluate the function as x approaches infinity. For the fractions, they approach 0.

(1 - 0)/(1 + 0)
1/1

1

Therefore, we have a horizontal asymptote at y = 1.

We will get the same result as x approaches negative infinity.

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Similarly, we can use long division; the quotient will end up being the horizontal asymptote.

(x^2 + 9) into (x^2 - 4) yields a quotient of 1 and a remainder of -13. That is,

(x^2 - 4)/(x^2 + 9) = -13/(x^2 + 9) + 1

The number not part of the fraction is the horizontal asymptote.

Answer 2

A, D, F that's a quadratic so has no maximum yet could have a minimum fee. once you positioned y to -4 and x to 2, the equation is balanced. once you positioned y to 0, x would be -2 or 3 (t locate the place it crosses one axis, set the different to 0).

Answer 3

D
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Reasons: There is no vertical asymptotes because the denominator cannot be zero. The horizontal asymptotes is found from its end behavior as x approaches infinity.

Answer 4

Only D

Answer 5

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