Question

Discontinuities of Rational Functions ALG. 2 -
What is the discontinuity & the zeros of:

f(x) = (4x^2 - 36x) ÷ (x – 9)

Answer 1

\[f(x)=\frac{4x^2-36x}{x-9}\] right?

Answer 2

yes.

Answer 3

\[x \neq 9\]

Answer 4

Yeah, but 9 is not a point on a graph, I also have to graph this. 9 is a restriction, not a point of discontinuity

Answer 5

it is certainly a point of discontinuity

Answer 6

you may not take
\[f(9)\] because it does not exist.

Answer 7

The equation is (4x^2 - 36x) ÷ (x - 9) How do you simplify it to 4(x^2 - 9)? What happened to the x of 36?

Answer 8

oh man i am sorry messed it up. it is
\[f(x)=\frac{4x(x-9)}{x-9}\] my mistake

Answer 9

ok it is still a point of discontinuity , because you cannot take
\[f(9)\]

Answer 10

this thing looks just like
\[y=4x\] except it has a hole at (9, 36)

Answer 11

the hole is the discontinuity

Answer 12

Yes I know that, thanks for clearing it up I kept trying (4, 9) or (4, 36)

Answer 13

ok as long as i didn't mess you up. just graph
\[y=4x\] but make a big hole at (9,36) and that is the picture

Answer 14

I posted the solution as well.

Answer 15

Thanks for ignoring me :S

Answer 16

I hate when I don't get proper credit for posting the solution

Answer 17

you said (3, -3).......

Answer 18

That's the solution for the zeroes, which is correct

Answer 19

I also said x cannot equal 9, which is also correct

Answer 20

The question asked not only for the discontinuity, but also the zeroes

Answer 21

Do you understand?

Answer 22

OK, sorry? You got your medal, and thanks for the help.

Answer 23

hold the phone

Answer 24

it is only 0 if x = 0, not at 3 or -3

Answer 25

Yeah, you're right, because it's linear. My bad

Answer 26

this is just
\[f(x)=4x,x\neq 9\]

Answer 27

yeah i factored incorrectly at first because i wasn't paying attention. i got that same answer!

Answer 28

can you explain how you got the zeros?

Answer 29

nevermind, i got it