Are there any asymptotes when you graph this function if so what is the equation of the asymptote?

Question

f(x)= x²- x-6

x²+ x-12

Answer 1

Hi Gina!

So to look for vertical asymptotes you look at where the denominator is 0. Since you can not divide by 0, that is why it’s undefined. So to find asymptotes you need to factor both the numerator and denominator.

f(x)= x²- x-6

x²+ x-12

f(x)= (x-3)(x+2) [ Multiplies to make -6 and adds to make -6]

(x-3)(x+4) [Multiplies to make -12 adds to make 1]

So if you notice there is (x-3) on top and bottom, that means there is a discontinuity, but that’s a hole, not an asymptote.

So to find the asymptote use the factor x+4, then set that equal to 0. So x+4 = 0 ——> x = -4 So the vertical asymptote is x=-4,

There is also a horizontal asymptote bc the degree of the numerator is the same as the degree of the denominator. So you find the leading terms of the top and bottom and it becomes a fraction: 1/1 therefore the horizontal asymptote is y=1

So asymptotes are x=-4 and y=1

I hope this helps! Have a great day!

Mrs. S

Answer 2

f(x) = (x - 3)(x + 2)/((x + 4)(x -3) = (x+2)/(x + 4)

Now we see that x = - 4 is vertical asymptote and y = 1/1 = 1 is horizontal asymptote and x = 3 f(x) has whole:

(3, 5/7)