Question

Find the equation of the tangent line to the curve y=((x-2)^3) * ((x-5)^2) at the point (3,4). Show work.

Answer 1

show work?

Answer 2

what defines a tangent?

Answer 3

I don't know this is for my girlfriend.

Answer 4

Derivative she says.

Answer 5

then you really cant participate in a meanginful conversation about this topic can you :/

Answer 6

But not sure.

Answer 7

derivative is good, but the derivative gives us the _____ at any given point along the curve

Answer 8

Nope. I appreciate your will to teach me, but its just that i'm studying for something else and my gf asked me for helo and i don't know this so i was like let me ask openstudy people

Answer 9

slope she says

Answer 10

things needed for an equation of a line are: a point and a slope

they give a point; so we need to determine the slope by using the derivative

Answer 11

the derivative of a product is:

[rl]' = r'l+rl'

Answer 12

Ok so she took a derivative and got y=8x-20

Answer 13

let me see if I can verify that :) i do need the practice at times lol

Answer 14

Hehe ok cool thanks =)

Answer 15

y=(x-2)^3 * (x-5)^2

y'=(x-2)'^3 * (x-5)^2 + (x-2)^3 * (x-5)'^2

y'=3(x-2)^2 * (x-5)^2 + (x-2)^3 * 2(x-5)

it might simplify down but im not concerned with that, I just need a number from it when I input our x component of the point

Answer 16

when x = 3 ...

y'=3(x-2)^2 * (x-5)^2 + (x-2)^3 * 2(x-5)

y'=3(3-2)^2 * (3-5)^2 + (3-2)^3 * 2(3-5)

y'=3(1)^2 * (-2)^2 + (1)^3 * 2(-2)

y'=3*4 + 1*-4

y'=12 -4 = 8

Answer 17

Oo ok I see ok. Thank you very very much! =)

Answer 18

so our slope at the given point is 8

y = 8x -8(3) + 4 would be one way to build the line

Answer 19

youre welcome :)

Answer 20

y=8x-20 is the end results, yes

Answer 21

Ok good perfect. Thanks again! =)