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: ?

Answer 1

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

Answer 2

m = f'(x)

Answer 3

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

Answer 4

f(x) = y = 4x^3

what would our derivative of this function?

Answer 5

Is it 4x^3?

Answer 6

thats the function itself; we want to take its derivative

Answer 7

\[kx^n\to knx^{n-1} \]

Answer 8

oh ok so 3x^4??

Answer 9

12x??

Answer 10

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 :)

Answer 11

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

Answer 12

our given point says x=-5

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

Answer 13

err, it says -4 doesnt it ..

Answer 14

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

Answer 15

yup

Answer 16

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

Answer 17

therefore b = -mPx + Py

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

yw, and good luck with it all :)

Answer 22

Thanks indeed!