ALGEBRA HELP PLEASE
, find the domain of each function

- Aizhalee

ALGEBRA HELP PLEASE
, find the domain of each function

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- Aizhalee

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- Aizhalee

@mathmate

- Aizhalee

thank you tkhunny for coming by to help me

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## More answers

- tkhunny

Please stop tagging people and start showing your work.
When is the denominator zero?

- Aizhalee

due to division it has to zero

- Aizhalee

f/g = 0

- tkhunny

Wherever the denominator is zero, we have a place that is NOT in the Domain. Where is that?

- Aizhalee

by setting each factor as zero ? to find x

- tkhunny

There you go. Factor it?

- Aizhalee

sorry I meant setting each factor EQUAL-ing to "0" x+1 = 0, x-12= 0. Am I right ? @tkhunny

- tkhunny

\(x^{2} + x - 12 = 0\)
Factor that. It's not 12 and 1.

- Aizhalee

OK (x+12) (x-1) ? Im sorry im just going by the way I have the notes in my notebook .

- Aizhalee

Thanks a lot Jim

- jim_thompson5910

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- Aizhalee

Yes I have that written down :) but how can I factor it correctly please

- jim_thompson5910

there is really a 1 in front of the x terms (if there aren't coefficients there already)
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- Aizhalee

yes sir

- jim_thompson5910

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- jim_thompson5910

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- jim_thompson5910

are you able to think of those two numbers?

- Aizhalee

6*-6 =-12

- jim_thompson5910

6 times -6 is not -12

- Aizhalee

my mistake 6*-2 =-12. I was thinking of additions >.<

- jim_thompson5910

that is true but 6 plus -2 does not equal +1 like we want

- jim_thompson5910

try another pair of factors

- Aizhalee

hmm ok 3*-4 ? or -1*12

- jim_thompson5910

3*-4 is very close
notice how 3 plus -4 = -1 but we want +1

- Aizhalee

yikes , ok um 2 + -1 = 1

- jim_thompson5910

you were on the right track, just you had to flip the signs
why not -3 and +4 ?
-3 times 4 = -12
-3 + 4 = +1
so `x^2 + x - 12` factors to `(x-3)(x+4)`
notice how the numbers -3 and 4 are part of the factorization like that

- Aizhalee

wow , Im sorry . and yes I do notice . thank you for explaining

- jim_thompson5910

so `x^2 + x - 12=0` turns into `(x-3)(x+4)=0`
what are the two values of x that make this equation true?

- Aizhalee

if (0)(7)=0

- jim_thompson5910

what did you replace x with to get that?

- Aizhalee

zeros oriel

- Aizhalee

huh?

- jim_thompson5910

how did you go from `(x-3)(x+4)=0` to `(0)(7)=0` ?

- Aizhalee

to add 3 to x

- jim_thompson5910

you replaced x with 3, so that shows that x = 3 is one solution

- jim_thompson5910

what is the other solution

- Aizhalee

3 honestly brcause 3-3 = 0 and 3+4=7

- jim_thompson5910

yes and because 0 times any number is 0

- jim_thompson5910

to get the other solution, solve x+4 = 0 for x

- Aizhalee

dont use x= 3 ?

- jim_thompson5910

no you already used that

- jim_thompson5910

you can subtract 4 from both sides of `x+4 = 0` to solve for x

- Aizhalee

x=-6

- jim_thompson5910

nope

- Aizhalee

6

- jim_thompson5910

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- jim_thompson5910

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- jim_thompson5910

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- jim_thompson5910

so hopefully you see that if `x+4 = 0` then `x = -4`

- Aizhalee

really? so 0-4 is -4 ?

- jim_thompson5910

yes

- Aizhalee

i thought the 0 would turn into 10 . but ok thank you for explaining jim

- jim_thompson5910

think of it like you're starting at 0 on the number line, then you move 4 spots to the left to end up at -4

- jim_thompson5910

The solutions to `x^2 + x - 12=0` are `x = 3 or x = -4`

- jim_thompson5910

the values `x = 3` and `x = -4` are the values that make the denominator zero, so you have to kick those values out of the domain
the domain would be the set of all real numbers BUT x cannot equal 3 or it cannot equal -4 (to avoid division by zero errors)

- Aizhalee

Ohhh. I get it now .Thank you so much Jim I am adding this in my nitebook .

- jim_thompson5910

np

- Aizhalee

Bye take care God bless :)

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