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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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