How do you find the asymptotes for #f(x)= (4x^2+5)/( x^2-1)#?

1 Answer
Mar 11, 2016

Answer:

To find vertical asymptote, you must set the denominator to 0 and then solve.

Explanation:

#x^2 - 1 = 0#

This can be factored as a difference of squares #(a^2 - b^ 2 = (a + b)(a - b))#

#(x + 1)(x - 1) = 0#

#x = -1 and 1 =># there are vertical asymptotes at x = -1 and x = 1.

Now for horizontal asymptotes, which are a little trickier.

To find these, you must look for the highest power (exponent) in both the numerator and the denominator. The highest power in both is #x^2#, which means that the asymptote will occur at the ratio of the coefficients of the highest power in the numerator and denominator. The coefficients are 4 and 1 respectively.

So, the horizontal asymptote occurs at #y = 4#

Verification:

Plugging in 4 for #f(x)#

#4 = (4x^2 + 5)/(x^2 - 1)#

#4(x^2 - 1) = 4x^2 + 5#

#4x^2 - 4 = 4x^2 + 5#

#0x^2 = 9#

#x^2 = 9/0#

#x = O/#, since division by 0 is undefined.

This proves that there is a horizontal asymptote at y = 4, because an asymptote is essentially and undefined line on the graph of a function.

Practice exercises:

  1. Identify all the asymptotes in #g(x) = (5x^2 + 10x)/(2x^2 + 7x + 3)#

Good luck!