What are the equations of the verticle,horizontal, and/or oblique asymtotes?

Question

f(x)= 2x+6 all over x-4

Answer 1

verticle: set the bottom part of the fraction equal to zero & solve for x... then those are your verticle asymtotes

so the verticle is: x-4 = 0 so x = 4 is the vertical asy.

horizontal asy.: look at all the x's... look at their exponents... if the top exponent is less than the bottom exponent, then the horizontal asy. is the x-axis. if the top exponent = the bottom exponent then you look at the numbers infront of each x (i'll explain that). if the top exponent is bigger than the bottom exponent, then there is no horizontal asy.

so the horiz. asy.: the x's are 2x (the top) and x (from the bottom)
each x has an exponent of 1... so you look at the number infront of each x and divide it. so you take 2/1 = 2 and that is your horizontal asy.

oblique asy.: if you can divide the bottom part of the fraction into the top part of the fraction by long division.... then that "quotient" is the equation of a line, which would be your oblique asymtote... since you have a verticle & horizontal asy., then you won't have an oblique asy.

Answer 2

f(x) = (2x + 6) / (x - 4)

To calculate the vertical asymptotes, you have to know what makes f(x) undefined. that is, x - 4 = 0, so x = 4 is a vertical asymptote.

To calculate the horizontal asymptotes, you have to find the limit as x approaches infinity.

lim (x --> infinity, (2x + 6)/(x-4))

First, divide all terms by the highest power (which is x)

lim (x --> infinity, (2 + 6/x)/(1 - 4/x) )

And for every term that has x as the denominator, it becomes 0 as x approaches infinity

= (2 + 0)/(1 - 0) = 2.

Therefore, y = 2 is a horizontal asymptote.

There aren't any oblique asymptotes.