Would i use the quotient rule first? ?

Remark.

I think your very first step is to find out exactly what f(x) is. If you don't do that, whatever you are given won't make any sense and you won't be able to go on.

Solution

Let f(x) = y

Go backwards.

y' = 3 when x = 2

dy/dx = 3

dy = 3 dx

y = 3x + C      Now you have to solve for C. y = 2 when x = 2

2 = 3*2 + C

C = - 4

f(x) = 3x - 4

g(x)

g(x) = x^2 * f(x)

g(x) = x^2 (3x - 4)

g(x) = 3x^3 - 4x^2

Tangent Line.

Slope

g'(x) is the slope of the tangent line

g'(x) = 9x^2 - 8x

g'(2) = 9(2^2) - 8*2

g'(2) = 36 - 16 = 20

So far what you have is y = 20x + b. Now you need to find the y intercept.

y intercept

g(x) = 3x^3 - 4x^2

g(2) = 3*2^3 - 4*(2^2)

g(2) = 3*8 - 4*4

g(2) = 24 - 16

g(2) = 8

What you have found is the point (2,8) which satisfies the line described by the slope about. Now you can find b

y = 20x + b

8 = 20*2 + b

8 = 40 + b

b = 8 - 40

b = -32

y = 20x - 32 is you final answer.

Comment

Though Part B is considerably longer, it is done the same way. I'm going to leave it for you. If you really need more help, just leave a note here.

For 13a, you use the product rule first to find the slope of the tangent, g'(2).

For 13b, you use the quotient rule first to find the slope of the tangent, h'(2).

13a. g'(x) = 2x·f(x) +x²·f'(x)

... g'(2) = 2·2·2 + 2²·3 = 20

... g(2) = 2²·2 = 8

The tangent line is y = 20(x -2) +8

13b. h'(x) = ((x -3)·f'(x) - f(x)·1)/(x -3)²

... h'(2) = (-1·3 -2)/(-1)² = -5

... h(2) = 2/(-1) = -2

The tangent line is y = -5(x -2) -2

The point-slope equation of a line is usually written (for slope m, point (h, k)) ...

... y -k = m(x -h)

For problems like this, I prefer the form with k added:

... y = m(x -h) +k

This saves a step if you decide you want to simplify it to slope-intercept form.

step-by-step explanation: 2