Finding the tangent and normal lines at x=2?

Question

Given f(x)=3x^3-5x+1 Find the equations of the tangent and normal (Perpendicular) lines to the curve at x=2. Write your answer in slope--y-intercept form.

Answer 1

x=2,
f(2) = 3(8) -10 +1 =15

f'(x) = 9x^2 -5
f'(2) = 36 -5 = 31 slope
eq. of tangent line: y-15 = 31(x-2)

eq. of normal line: y-15 = (-1/31) (x-2)

Answer 2

1) find the point at x=2

f(x)=3x^3 +5x +1, then f(2)=3(8)+5(2)+1 = 35

2) find the slope of the tangent. First differentiation:
f'(x) = 9x^2 + 5, therefore f'(2) = 39

3) write the tangent in slope-intercept form:

y = mx + c

From 1, we have values for y and x (35 and 2).
From 2, we have value for m (39)
find c.

4) Perpendicular: slope of perpendicular is 1/m (1/39)

y = (1/m)x + p (where p is the y-intercept of the perpendicular)

From 1, we have values for y and x
From 2, we have value for 1/m
Find p.

Answer 3

f'(x) = 3(3)x^2 - 5 = 9x^2 - 5
This is the slope of the line at x, f'(x=2) = 9(2)^2 - 5 = 31
The slope intercept form
y(x) = mx + b where m is slope, in this case m = 31
y(x) = (31)(2) + b = f(2) = 3(2)^3 - 5(2) + 1 = 15
so that b = 15 - 62 = -47
y(x) = 31x - 47 is the eqn of the tangent

The normal: do it pretty much the same way keeping in mind that the slope of the normal is the negative inverse of the slope of the tangent line:
Mnormal = -1/31
Ynormal(x) = Mnormal(x) + B
(-1/31)x + B