How do you find all the asymptotes for function #y=(3x^2+2x-1)/(x^2-4 )#?

Question

Ariana Alvarez · Accepted Answer

Vertical asymptotes are #x+2=0# and #x-2=0#. Horizontal asymptote is given by #y=3#

To find all the asymptotes for function #y=(3x^2+2x−1)/(x^2−4)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2-4=0# i.e. #x+2=0# and #x-2=0#.

As the highest degree of both numerator and denominator is #2# and ratio of these is #3x^2/x^2# i.e. #3#, horizontal asymptote is given by #y=3#.

Vertical asymptotes are #x+2=0# and #x-2=0#. Horizontal asymptote is given by #y=3#

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I hope you do not mind but I added a graph.( Tony B)

Ariana Alvarez · Answer

undefined To find the asymptotes of the function $y = \frac{3x^2 + 2x - 1}{x^2 - 4}$, we need to look at both the vertical and horizontal asymptotes.

**Vertical Asymptotes:** Vertical asymptotes occur where the denominator of a rational function is equal to zero, but the numerator is not zero. In this case, the denominator $x^2 - 4$ is equal to zero when $x = 2$ or $x = -2$. However, at these points, the numerator $3x^2 + 2x - 1$ is not equal to zero. Therefore, the vertical asymptotes are $x = 2$ and $x = -2$.

**Horizontal Asymptotes:** Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. In this case, the degree of the numerator is 2 and the degree of the denominator is also 2. To find the horizontal asymptote, divide the leading terms of the numerator and denominator. The horizontal asymptote is the ratio of the leading coefficients. So, the horizontal asymptote is $y = \frac{3}{1}$ or simply $y = 3$.

Therefore, the function $y = \frac{3x^2 + 2x - 1}{x^2 - 4}$ has vertical asymptotes at $x = 2$ and $x = -2$, and a horizontal asymptote at $y = 3$.