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: ?

Question

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:  ?

amistre64 · Answer

the derivative of a function defines an equation to calculate the slope at any point along the cirve

amistre64 · Answer

m = f'(x)

amistre64 · Answer

b = -f'(x)x+f(x)

amistre64 · Answer

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

anonymous · Answer

Is it 4x^3?

amistre64 · Answer

thats the function itself; we want to take its derivative

amistre64 · Answer

$$kx^n\to knx^{n-1} $$

anonymous · Answer

oh ok so 3x^4??

anonymous · Answer

12x??

amistre64 · Answer

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 :)

amistre64 · Answer

12x^2 .... youre getting there :)

amistre64 · Answer

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

amistre64 · Answer

err, it says -4 doesnt it ..

amistre64 · Answer

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

anonymous · Answer

yup

amistre64 · Answer

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

amistre64 · Answer

therefore b = -mPx + Py

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

Oh ok wow! Thanks for your great help amistret64!!

amistre64 · Answer

yw, and good luck with it all :)

anonymous · Answer

Thanks indeed!