Find the tangent line of y=7x^3 + 2x at x=-3?

Question

I know that y-f(c)=f'(c)(x-c) but how do I use this to get the tangent line, in standard form , to y=f(x) at the indicated point? Please explain step by step with as much detail as possible because I'm completely lost.

Answer 1

First find the y-coordinate of the point,
y(-3) = 7(-27) - 6
y(-3) = -195
point is (-3, -195)

slope = dy/dx = 21x^2 + 2
y'(-3) = 191

Eq. of the tangent line is (y - y1) = m(x - x1),

y - (-195) = 191(x - (-3))

y + 195 = 191x + 573

y - 191x - 378 = 0

Answer 2

Find the derivative. dy/dx = 21x^2 + 2
or f`(c)
Plug in your point. You get 21(-3)^2 + 2 = f`(-3)
This will give you the slope of the line at -3 of 191.

Remember the point slope form of y-y1 = m(x-x1)
(the same as the equation you wrote)
In this case x1 = -3, and y1 = 7*-3^3 + 2 * -3 or y1 = -195
So your answer is y - 191x - 378 = 0

Answer 3

Take the derivative of y to get y'(x)=21x^2+2
Evaluate the derivative at x=-3 to get y'(-3)=191 which is equal to the slope dy/dx
Evaluate y(-3)=-195
Now you know the slope (191) at a point (-3,-195)
Use m = (y-y1)/(x-x1) where m is the slope and (x1,y1)=(-3,-195) and express y=f(x)

Answer 4

easiest way to do this by far is to just throw it into your graphing calculator. i dont know what kind you have, but if you have a TI 83 or 84, put in the equation in y=, then push graph, then go 2nd prgm, then tangent, then put in -3, and push enter. you get y=191x+378