Question

find the verical asymptotes if any and the values if any of (x) and the corresponding holes if any of the rational graph f(x)= x^2-16/x-4

Answer 1

well we have a fraction
the fraction is undefined when the bottom is 0
when is the bottom 0?

Answer 2

because x-4 is x+4

Answer 3

which = 0

Answer 4

x-4 isn't x+4

i'm asking when the bottom is 0
when is x-4=0

Answer 5

then I dont know

Answer 6

you never solve equations like x-4=0 before?

Answer 7

i will give you a hint
add 4 on both sides

Answer 8

x-4+4=0+4

Answer 9

what is -4+4?

Answer 10

0

Answer 11

so x+0=0+4
so x=4
x-4=0 when x=4
the bottom is 0 when x=4

Answer 12

that means there is a hole at x=4 or a vertical asymptote

Answer 13

to figure out which one we need to factor the top and see if we can get the bottom factor to cancel out

Answer 14

can you factor x^2-16?

Answer 15

ok but what about the x^2-16

Answer 16

i know that 4 goes in to 16 4 times

Answer 17

but the x^2 has me stumpt

Answer 18

x^2-16 is a difference of squares

to factor a^2-b^2 we get (a-b)(a+b)

Answer 19

ok

Answer 20

i hate these i never end up understanding a dam thing

Answer 21

well right now i'm just asking you to factor x^2-16

Answer 22

you can use the formula i gave if you want

Answer 23

that formula is were i always can no figure things out

Answer 24

x=0 (x-16)(x+16) ?

Answer 25

x^2-16=x^2-4^2=(x-4)(x+4)

Answer 26

\[f(x)=\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}\]
if we are able to cancel the x-4 on bottom then we have a hole at x=4
but if we we can't cancel the x-4 on bottom then we have a vertical asymptote at x=4

but as you notice it does cancel