Need help finding the tangent line with a given point that is NOT on the stated curve

Q: http://imageshack.us/photo/my-images/713…

the point given isnt on the curve, so i only got to what the slope is from taking the derivative, but what do i do after that?

any tips or hints or answers will be GREATLY appreciated
7 minutes ago - 4 days left to answer.

You are looking for a line, or lines, through (1,-3) that are tangent to the curve.
y = 4x³-15x + 12

Slope: y' = 12x²-15
Line: (y+3) = y'(x-1)
y = y'x - y'-3
y = y'x - (y'+3)

Substituting for y and y':
4x³-15x+12 = (12²x-15)x - (12x²-15+3)
4x³-15x+12 = 12x³ - 15x - 12x² + 12
12x³-4x³ -12x² - 15x+15x+12-12 = 0
8x³-12x² = 0
x²(8x-12) = 0
x = 0, 0, 3/2

f(0) = 12
f(3/2) = 4(3/2)³-15(3/2) + 12 = 4*27/8 - 45/2 + 12 = (27-45+24)/2 = 3

There are two lines through the point (1,-3) that are tangent to the curve:
Line 1: Tangent point (0, 12), Slope = 12x²-15 = -15
(y-12)/x = (-3-12)/(1-0) = -15

y = -15x + 12
or
15x + y = 12

Line 2: Tangent point (3/2, 3), Slope = 12(3/2)² - 15 = 12
(y-3)/(x-3/2) = (-3-3)/(1-3/2) = 12

y = 12x-15
or
12x - y = 15

TRivial.
you are given a point (a,b) and a function defining a curve f(x)
so you know an equation for the tangent line at every point on the curve f'(x)
so you also know (or can easily figure) the slope for the line connecting (x,f(x)) to (a,b)
when that slope equals the tangent line at the point x, then obviously that point (x,f(x)) will be the point(s) whose tangent line intersects (a,b)
slope [x,f(x)] to (a,b) equals slope at point x = f'(x)= d(f(x))/dx