Find the equation of the tangent line to the curve y=((x-2)^3) * ((x-5)^2) at the point (3,4). Show work.
what defines a tangent?
I don't know this is for my girlfriend.
Derivative she says.
then you really cant participate in a meanginful conversation about this topic can you :/
But not sure.
derivative is good, but the derivative gives us the _____ at any given point along the curve
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
slope she says
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
the derivative of a product is: [rl]' = r'l+rl'
Ok so she took a derivative and got y=8x-20
let me see if I can verify that :) i do need the practice at times lol
Hehe ok cool thanks =)
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
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
Oo ok I see ok. Thank you very very much! =)
so our slope at the given point is 8 y = 8x -8(3) + 4 would be one way to build the line
youre welcome :)
y=8x-20 is the end results, yes
Ok good perfect. Thanks again! =)
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