How do you find vertical, horizontal and oblique asymptotes for #f(x)= (5x^2-17x-12)/(2x^2-7x-4)#?

1 Answer
Oct 23, 2016

Answer:

The horizontal asymptote is #y=5/2#, the vertical asymptote is #x=-1/2# and there is no oblique asymptote.

Explanation:

#f(x)=frac{5x^2-17x-12}{2x^2-7x-4}#

This problem is a bit tricky because there is a "hole" at #x=4#.

Start by factoring both numerator and denominator.

#f(x)=frac{(5x+3)(x-4)}{(2x+1)(x-4)}#

Note that factor #(x-4)# cancels. This means there is a "hole" or undefined point in the graph at #x=4#.

To find the asymptotes, use the function that results after the factor cancels.

#f(x)=frac{5x+3}{2x+1}#

To find the horizontal asymptote (HA), compare the degree of the numerator to the degree of the denominator.

If the degrees are the same, the HA is #y=# leading coefficient of the numerator divided by the leading coefficient of the denominator.

If the degree of the denominator is greater, the HA is #y=0#.

If the degree of the numerator is greater, there is an oblique asymptote.

#f(x)=frac{color(blue)5x^color(red)1+3}{color(blue)2x^color(red)1+1}#

In this example, both the numerator and denominator have a degree of #color(red)1#. (The degree is the highest exponent in standard form).

The HA is then #y=color(blue)5/color(blue)2#

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.

To find the vertical asymptote (VA), find the value of #x# where the denominator equals zero. This value of #x# will result in dividing by zero, which is undefined.

#2x+1=0 color(white)(aaa)#Solve for x

#2x=-1#

#x=- 1/2color(white)(aaa)#This is the equation of the VA.