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OpenStudy (s):

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

OpenStudy (amistre64):

show work?

OpenStudy (amistre64):

what defines a tangent?

OpenStudy (s):

I don't know this is for my girlfriend.

OpenStudy (s):

Derivative she says.

OpenStudy (amistre64):

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

OpenStudy (s):

But not sure.

OpenStudy (amistre64):

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

OpenStudy (s):

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

OpenStudy (s):

slope she says

OpenStudy (amistre64):

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

OpenStudy (amistre64):

the derivative of a product is: [rl]' = r'l+rl'

OpenStudy (s):

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

OpenStudy (amistre64):

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

OpenStudy (s):

Hehe ok cool thanks =)

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (s):

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

OpenStudy (amistre64):

so our slope at the given point is 8 y = 8x -8(3) + 4 would be one way to build the line

OpenStudy (amistre64):

youre welcome :)

OpenStudy (amistre64):

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

OpenStudy (s):

Ok good perfect. Thanks again! =)

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