Question

for the normal (perpendicular) line to the curve y= square root of 8-x^2 at (-2,2) would the slope be 1/2?

Answer 1

what is f'(-2) ?

Answer 2

f(x) = 8-x^2

f'(x) = -2x

Answer 3

the derivative gives us the tangent slope at any given point

Answer 4

a perp slope (the normal) is the negative reciprocal

Answer 5

\[-\frac{1}{-2x}=\text{normal slope}\]
when x=-2 we get -1/4 if i did it right

Answer 6

well the choices are -2, 1/2 , -1/2 , 1 , -1 i thought it was 1/2 because when i did the work i ended it up with what u had but i thought the negative signs would cancel eachother out.

Answer 7

3 negatives is still negative :/

Answer 8

opps, i didnt read the square root of part

Answer 9

i didnt see the third negative grrrrr. -_____- i was so close ! :[

Answer 10

lol and to think i thought i did this right i felt proud for a second.

Answer 11

sqrt(8-x^2) is easier to see mathically ;)

Answer 12

you did good, those signs still fool me too

Answer 13

hmm so it is -1/2 ? aww I was so close.

Answer 14

\[[\sqrt{8-x^2}]'=\frac{-2x}{2\sqrt{8-x^2}}\]
\[-\frac{\sqrt{8-x^2}}{-x}\]
\[-\frac{\sqrt{8-(-2)^2}}{-(-2)}\]
\[-\frac{\sqrt{8-4}}{2}\]
\[-\frac{2}{2}=-1\]
so did I do it right?

Answer 15

it might be easier to solve for the f' and then flip and negate it:

\[\frac{-x}{\sqrt{8-x^2}}\]
\[\frac{-(-2)}{\sqrt{8-(-2)^2}}\]
\[\frac{2}{\sqrt{8-4}}\]
\[\frac{2}{\sqrt{4}}=\frac{2}{2}=1\]
flip and negate to -1

Answer 16

hmmm I see what you did, ehh calculus <.< I feel like a scrub now lol.

Answer 17

;) what were you trying?

Answer 18

I started in the right the right direction, but after finding the derivative idk what I did to be honest lol.

Answer 19

keep at it, youll do fine, good luck!

Answer 20

Haha thanks :DD