Question

Which of the following represents the vertical asymptotes of the function f(x) =(3x - 8)/(x^2 - 3x - 10)?

Answer 1

x = 2 and x = 5

x = –2 and x = 5

x = 2 and x = –5

x = –2 and x = –5

Answer 2

verts are what makes the denom go zero, but that do not have a canceling factor in the numerator

Answer 3

it looks like the options are simple enough; just plug them into the denominator to see what zeros it out

Answer 4

i dont understand how to do it though. I'm not good with asymptotes and I don't know what they mean when they say they have to be vertical.

Answer 5

do I have to graph it?
would that help?

Answer 6

do you know how to factorise the denominator?

Answer 7

would it be (x - 5)(x + 2)?

Answer 8

yes

Answer 9

now see what valus of x would make this equal to zero

Answer 10

i.e. what values of x satisfy this equation:\[(x-5)(x+2)=0\]

Answer 11

make sure theres no canceling factor in the numerator

Answer 12

yes ^^

Answer 13

so is the answer x = -2 and x = 5 ?

Answer 14

what we have done so far is:\[\frac{3x-8}{x^2-3x-10}=\frac{3x-8}{(x-5)(x+2)}\]do you think there are any common factors between the numerator and denominator here?

Answer 15

yes - your answer is correct
what @amistre64 was trying to show is that before the step of finding what values of x make (x-5)(x+2)=0, it is important to cancel out any common factors between the numerator and denominator.

Answer 16

ohhh ok thanks

Answer 17

e.g., lets say your equation was actually:\[\frac{x+2}{(x-5)(x+2)}\]

Answer 18

then we would first cancel out the (x+2) terms

Answer 19

leaving 1/(x-5)

Answer 20

so the only asymptote here would be x=5