Question

how do i graph f(x) = (9x^2 - 36) / (3x + 6)

@misty1212

Answer 1

ok so we have this->\(\huge\ f(x)= \frac{ 9x^2-36 }{ 3x+6 }\)


try to apply this identity in the numerator->\(\large\color{violet}{ a^2-b^2=(a+b)(a-b)}\)

Answer 2

you end up graphing \(y=3x-6\) a line

Answer 3

9(x+2)(x-2)

Answer 4

what did you do

Answer 5

difference of squares

Answer 6

ok thats not correct

see it like this->\(\huge\ f(x)= \frac{ (3x)^2-(6)^2 }{ 3x+6 }\)
now try :)

Answer 7

factor out using GCF?

Answer 8

no
1st you apply this identity in the numerator and factor the terms-\(\large\color{violet}{ a^2-b^2=(a+b)(a-b)}\)

and then you cancel out the common factors in numerator and denominator

Answer 9

it is the same thing, makes no difference

Answer 10

oh sorry i didn't noticed >.<

Answer 11

you can factor as \(9(x^2-4)=9(x+2)(x-2)\)

Answer 12

yea thats what i got.. and then ?

Answer 13

and denominator as \[3(x+2)\]

Answer 14

then cancel and get \[3(x-2)\]

Answer 15

or \[y=3x-6\] same thing either way

Answer 16

ok and what number will be circled when graphing..

Answer 17

if \(x+2=0\) then \(x=-2\)

Answer 18

so you can't put in \(-2\) that will have the circle

Answer 19

second coordinate will be \[3(-2)-6=-12\]

Answer 20

these are my choices.. i know its not A or B..

Answer 21

go with C as usual

Answer 22

the one with the circle at \((-2,-12)\)

Answer 23

ok but how come ?

Answer 24

you cancelled the \(x+2\) from the denominator , which is fine, but the domain does not include \(-2\) since the denominator would be zero there

Answer 25

if you put \(-2\) in the cancelled expression \(3x-6\) you get \(-12\)

Answer 26

oh....... okok let me do another one to make sure.. ill tag you : )