Question

Find the limit of the function algebraically.
limx->-6 x2 -36/x + 6

Answer 1

@satellite73

Answer 2

factor
cancel
replace \(x\) by \(-6\)

Answer 3

i can walk you through it if you like , but those are the steps

Answer 4

Okay how do I factor? I can do the rest. I think.

Answer 5

factor
\[x^2-36=(x+6)(x-6)\]

Answer 6

\(\Large\frac{x^2-36}{x+6}=\frac{(x+6)(x-6)}{(x+6)}\)

Answer 7

i.e. the difference of two squares

Answer 8

then cancel, get \(x-6\) then replace \(x\) by \(-6\)

Answer 9

I think there's a -6 though as the coefficient of x^2. At least that's how I am interpreting what she originally wrote.

Answer 10

No, It is just asking for the one sided limit coming for the left at x=-6

Answer 11

Ahh I see okay, thanks for the clarification. My mistake.

Answer 12

here is all the steps, written in one line
\[\lim_{x\to -6}\frac{x^2-36}{x+6}=\lim_{x\to -6}\frac{(x+6)(x-6)}{x+6}\\
=\lim_{x\to -6}x-6=-6-6=-12\]

Answer 13

Aahhhh thank you for the clarification

Answer 14

it is not a "one sided limit" it is just the limit
hope the steps are clear, very many work exactly this way

Answer 15

Oh yeah sorry I also read it wrong because of the -> nice work @satellite73

Answer 16

you want to try another or are you done?

Answer 17

Ummm can you help with this one?

Find the center, vertices, and foci of the ellipse with equation 5x^2 + 8y^2 = 40.

Answer 18

sure but this is totally different

Answer 19

I know. Sorry.

Answer 20

it is not that hard though

start by dividing by 40 to get
\[\frac{x^2}{8}+\frac{y^2}{5}=1\]

Answer 21

Okay what's next?

Answer 22

then you are really almost done
is it clear that the centre is \((0,0)\) ?

Answer 23

Yes I see that

Answer 24

ok and it is also clear that the larger number is under the \(x^2\) term, so it looks like this

Answer 25

Okay I'm with you so far

Answer 26

general form is
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] if the center is \((0,0)\)
that makes \(a^2=8\) and so \(a=\sqrt8=2\sqrt2\)

Answer 27

you need to know what it looks like to see that the foci are \(a\) units to the RIGHT and LEFT of the center
so the foci are \((-2\sqrt2,0)\) and \((2\sqrt2,0)\)

Answer 28

whoa scratch that!!!

Answer 29

all of what i said there was true, but it is for the VERTICES not the foci
my mistake

Answer 30

vertices are \((-2\sqrt2,0)\) and \((2\sqrt2,0)\)

Answer 31

damn i am full of typos
let me know if you are confused
if not we can go on and find the foci
it is pretty easy really

Answer 32

Okay so the foci are \[(-\sqrt{3} ,0) (\sqrt{3}, 0)\]

Answer 33

yup

Answer 34

Can we do another?

Answer 35

lol sure

Answer 36

hope i can go easier on the typos

Answer 37

Find the vertices and foci of the hyperbola with equation (x + 2)^2/144 - (y - 4)^2/81= 1

Answer 38

Sorry, I realized I never hit post

Answer 39

k at least this one is in the right form to start
you got the center?

Answer 40

\[\frac{(x+2)^2}{12^2}-\frac{(y-1)^2}{9^2}=1\]

Answer 41

Okay so the center is (0,0)

Answer 42

right?

Answer 43

oh no

Answer 44

oh...I'm confused

Answer 45

\[\frac{(x+2)^2}{12^2}-\frac{(y-1)^2}{9^2}=1\]
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] center is \((h,k)\)

Answer 46

they have shifted
it is not \(x^2\) and \(y^2\) up top
your center for \[\frac{(x+2)^2}{12^2}-\frac{(y-1)^2}{9^2}=1\] is \((-2,1)\)

Answer 47

ohhhhhh okay

Answer 48

ok we need that first
we also need to know what it looks like