Question

Let f be the function (square root(x^4-16x^2))

find the slope of the line normal to the graph of f at x=6.

Answer 1

the slope is the derivative; what do we find as the derivative?

Answer 2

i already got the deriv it wud be (2x^2-16x)/ square root of (x^4-16x^2) i just dont know what it means when they ask lline normal to i didnt learn that yet

Answer 3

that is the deriv right? ><

Answer 4

4x^3 - 32x
--------------- is what i get; then when x=6 .....
2sqrt(x^4-16x^2)

ohh, i read slope of the line; the normal is the perp slope; 1/f'(x) in this case

Answer 5

in other words; take f'(6) and flip it

Answer 6

if a tangent slope is: 4/5

the normal slope for that would be : 5/4

Answer 7

okay hang on gona do that ^^ nd thank you for helping

Answer 8

forgot a negative

Answer 9

if a tangent slope is: 4/5

the normal slope for that would be : -5/4


thats better

Answer 10

oh okay so its kinda like when your doing slopes perpendicular to?

Answer 11

yes

Answer 12

not kinda; its the exact same thing :)

Answer 13

oh they're just out to get us with different wording >< GAH okay so i know how to do it thank you so much ^^

Answer 14

youre welcome :)

Answer 15

^^

Answer 16

uhm sorry again, but i got a weird fraction >< idk if my answer is correct can you just check it please?

Answer 17

4x^3 - 32x
---------------
2sqrt(x^4-16x^2)



2x^3 - 16x
------------------
sqrt(x^2)srt(x^2-16)



2x^3 - 16x
-----------
xsqrt(x^2-16)



2x^2 - 16
----------- ; x=6 we get .... 56/sqrt(20)
sqrt(x^2-16)


56/2sqrt(5) is the slope of the tangent

-sqrt(5)/28 should be the slope of the normal

Answer 18

hmmm, wolf says i mighta had an error

Answer 19

uhm.. hang on checking with mine

Answer 20

2(36)-16 = 72-16 = 62-6 = 56
sqrt(6^2-16) = sqrt(36-16) = sqrt(20) = 2sqrt(5)

56/2 = 28

28/sqrt(5) is what i get for f'(6) still

Answer 21

http://www.wolframalpha.com/input/?i=derivative+sqrt%28x%5E4-16x%5E2%29%3B+x%3D6

now it likes it ....

Answer 22

yeah, im good again lol

Answer 23

wait i think i might have accidently squared a number when i shudnt have /fail hang on again >< really sorry T.T

Answer 24

for my f'(x) i got 2x^3-16x / square root (x^4-16x)

Answer 25

i dont get how u got 4x^3 - 32x
---------------
2sqrt(x^4-16x^2)