How do you identify the vertical and horizontal asymptotes, and any holes in the graph?

Question

f(x) = (x^2 + 3x - 18) / (x^2- 36)

Answer 1

For the vertical asymptotes, just determine where the denominator is equal to zero ( also check to see if the numerator is zero at this same value--because this might be a removable singularity). For the horizontal asymptote, divide the coefficient of the leading term of the numerator by the leading term of the denominator: in this case 1/1 =1. So y=1 is the horizontal asymptote. Note that if the power of the top leading term was greater then the bottom leading term, their would be no horizontal asymptote.

Answer 2

Factor the numerator and denominator first.
f(x)=[(x+6)(x-3)]/[(x+6)(x-6)]
Cancel the factors (x+6). This is what will create the hole, but first rewrite the function.
f(x)=(x-3)/(x-6)
Find the hole by taking the factor x+6 and setting it equal to 0.
x+6=0
x=-6, put this into the rewritten function to find the y coordinate of the hole.
f(-6)=(-6-3)/(-6-6)
f(-6)=3/4, the hole is (-6, 3/4)
For the vertical asymptote take the denominator from the new function, f(x)=(x-3)/(x-6), and set it equal to 0.
x-6=0
So Vertical Asymptote is x=6
Horizontal asymptotes can be tricky. Since the degree of the numerator is the same as the degree of the denominator the horizontal asymptote is created by the coefficients in front of the x's. They're both 1 so y=1 is the horizontal asymptote.

Answer 3

hi first y=f(x) = (8 x + 2) / (3 x + 3) =(8/3)* (x + a million/4)/(x+a million) = (8/3)*[ a million - (3/4)/(x+a million)] horizontal asymptotes : x = -a million x + a million = (3/4) * a million/ ( a million - (3/8)y) vertical asymptotes : y = 8/3 2d: y = f(x) = (2 x^2 - 8) / (3 x^2 + 9 x - 30) = (2/3) * (x^2 - 4) / (x^2 + 3 x - 10) = (2/3) * [(x+2)(x-2)/(x+5)(x-2)] for x ne 2 = (2/3) * [(x+2)/(x+5)] = (2/3) * ( a million - 3/(x+5)) no necessary discontinuity x = 2 in ( 2 ; 8/21) horizontal asymptotes : x = -5 x + 5 = 3 / ( a million - (3/2) y) vertical asymptotes : y = 2/3 bye

Answer 4

First, factor each part. If one factor "cancels" out, do so, but rememeber that there will be a hole where that factor would have been zero.

Then to find vertical asymptotes, find what makes the (remaining) bottom be zero.

To find the horizontal one, when the degrees are equal, divide the leading coefficients. So it is y = 1/1

Answer 5

f(x) = (x^2 + 3x - 18)/(x^2 - 36)
f(x) = ((x + 6)(x - 3))/((x - 6)(x + 6))
f(x) = (x - 3)/(x - 6)

x cannot equal 6 or -6

f(x) = ((x + 6)(x - 3))/((x - 6)(x + 6))
y = (x - 3)/(x + 6)

to find out what "y" cannot be, just find the inverse of f(x)
x = (y - 3)/(y - 6)
xy - 6x = y - 3
xy - y = 6x - 3
y(x - 1) = 6x - 3
y = (3(2x - 1))/(x - 1)

y cannot equal 1

ANS :
Horiztonal Asymptote is at y = 1
Vertical Asymptote is at x = 6 and x = -6

The hole in this case is at x = -6 because (x + 6) is the only one that cancelled out.