Question

@ganeshie8

Answer 1

The vertical asymptotes(s),if any, would be:


Select one:


a. x = 2

b. x = 2 and x = –3

c. y = 2 and y = –3

d. No vertical asymptotes

Answer 2

ow they are asking for vertical asymptotes

Answer 3

factor the numerator
factor the denominator

Answer 4

familiar with factoring ? :)

Answer 5

kind of

Answer 6

we need to cancel common factors, if any, first

Answer 7

for that we need to factor top and bottom

Answer 8

just like last time?

Answer 9

nope

Answer 10

4x^2+12x = 4x(x+3)

yes ?

Answer 11

can you factor denominator like this ?

Answer 12

x^2+x-6 = (x+?)(x-?)

Answer 13

um let me see

Answer 14

how do I do that though it makes no sense:(

Answer 15

x^2+x-6 = (x+3)(x-2)

Answer 16

\[\large \begin{align} y &= \dfrac{4x^2+12x}{x^2+x-6}\\~\\&= \dfrac{4x(x+3)}{(x+3)(x-2)}\\~\\ \end{align}\]

Answer 17

I'll explain you how to factor some other time :)
for now just see that (x+3) cancels out in both numerator and denominator

Answer 18

so for the answer we are looking at B or C correct

Answer 19

\[\large \begin{align} y &= \dfrac{4x^2+12x}{x^2+x-6}\\~\\&= \dfrac{4x(x+3)}{(x+3)(x-2)}\\~\\ &= \dfrac{4x}{x-2}\end{align}\]

Answer 20

For vertical asymptote, set the DENOMINATOR equal to 0

(memorize this by heart)

Answer 21

\[\large x -2 = 0 \]

Answer 22

ohh so the answer would be A x=2

Answer 23

Exactly !

Answer 24

only one vertical asymptote we got

Answer 25

cuz the other factor in the denominator eaten by numerator

Answer 26

and are you a boy or girl