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howabout x^2 + 2 AND x+8

Drawing a sketch helps.

How did you determine those intervals?

Does that make sense?

absolutely. Thank you.

Certainly.

take the derivative of the function to find the slope. then plug it into slope intercept form?

Yes.

\[y-y_1 = m(x-x_1)\]

Yes that one, sorry I was confused.

However, when I took the derivative, it turned out complicated. Can you run through that with me?

Can you take the derivative of that?

yes, by chain rule

What do you get for the derivative

Actually I'm thinking that we might want the derivative of \[-x\sqrt{x+7}\]

why did it become negative?

hey polpak the answer to my question was D right

Well there are two options.
\[y^2 = x^2(x+7) \implies y = \pm\sqrt{x^2(x+7)} = \pm x\sqrt{x+7}\]

And I'm not sure how to tell which one we want. I think it's the negative one.

yes! I'm sure it's the negative one

Because y = 6, but x = -3. So that means that -x = 3

\[f'(x) = \sqrt{x + 7} + {x}\ {2 sqrt {x + 7}}]

Close. It should be \[\sqrt{x+7} + \frac{x}{2\sqrt{x+7}}\]

Oh, except negative.

-1 times that whole thing.

That exacty what I got.

So now plug in -3 for x.

how did you get -2x - 14 for your numerator?

For getting everything over the same denominator.

Gives
\[\frac{-2(x+7)}{2\sqrt{x+7}}\]

allright now what?

Now plug in -3 for x and find your slope at that point.

I got y = -5/4x - 9/2

y-6 = -5/4(x+3)
y = (-5/4)x -15/4 + 24/4
y = (-5/4)x - 9/4

Not -9/2

Did you use point slope?

Or did you use slope intercept?

slope intercept. my points were (-3, 6) and I plugges that into y2-y1= m (x2-x1)

Just use point slope.

I know where I went wrong, i reversed my x and y values

How do I find the equation of a normal line now that I found the equation of a tangent line?

Well the normal line is at that same point, but the slope is perpendicular.

Perpendicular to the tangent line that is

4/5 is my slope?

Yes, the slope of the normal line is 4/5

You don't need to. The normal at that point still touches that point (-3,6)

You just need one point and the slope to find the equation of the line.

for where the tangent line is vertical?

0 in the denominator of the derivative that is.

Help? :/

What is the denominator of the derivative?

oh, set the denominator to zero?>

oh, set the denominator to zero?>

x is greater than or equal to -7

Can I bother you with one more problem?

yes, but go ahead and make a new question for it. this one is getting long ;)