Mathematics OpenStudy (anonymous):

The slope of the tangent line to the curve y=4x^3 at the point (-4, -256) is:? The equation of this tangent line can be written in the form y=mx+b where m is: ? and where b is: ? OpenStudy (amistre64):

the derivative of a function defines an equation to calculate the slope at any point along the cirve OpenStudy (amistre64):

m = f'(x) OpenStudy (amistre64):

b = -f'(x)x+f(x) OpenStudy (amistre64):

f(x) = y = 4x^3 what would our derivative of this function? OpenStudy (anonymous):

Is it 4x^3? OpenStudy (amistre64):

thats the function itself; we want to take its derivative OpenStudy (amistre64):

$kx^n\to knx^{n-1}$ OpenStudy (anonymous):

oh ok so 3x^4?? OpenStudy (anonymous):

12x?? OpenStudy (amistre64):

I would advice you look over the basic rules for derivatives so that you can get a handle on them; in the mean time: k = 4 and n=3 .... can be plugged into the general setup I offered you as a reminder :) OpenStudy (amistre64):

12x^2 .... youre getting there :) OpenStudy (amistre64):

our given point says x=-5 so our slope is defined as f'(-5) = 12(-5)^2 OpenStudy (amistre64):

err, it says -4 doesnt it .. OpenStudy (amistre64):

f'(-4) = 12(-4)^2 would be more appropriate in this case then OpenStudy (anonymous):

yup OpenStudy (amistre64):

the b parts we can establish from moving about the point slope form of the equation of a line: given a slope m and a point (Px,Py) y-Py = m(x-Px) y-Py = mx-mPx y = mx-mPx + Py OpenStudy (amistre64):

therefore b = -mPx + Py OpenStudy (anonymous):

Oh ok I understand how you get the first one now...:) OpenStudy (amistre64):

$b = [-12(-4)^2\times -4] + (-256)$ looks about right :) OpenStudy (anonymous):

Oh ok wow! Thanks for your great help amistret64!! OpenStudy (amistre64):

yw, and good luck with it all :) OpenStudy (anonymous):

Thanks indeed!

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