Find all the points on the curve y=3x^2 where the tangent line passes through (2,9)?

Question

So I was trying to use the derivative of y=3x^2, which is y'=6x. Plug x=2 into that so the slope at x=2 is 12. I actually have no clue if that's right or not. Anyway, I'm pretty clueless after that anyway... thanks for the help in advance.

Oh that makes sense. Thanks!

Answer 1

1st part is ok, slope of tangent is d(3x²)/dx = 6x, but (2,9) is not a point on the curve. you have to find a point on the curve, (x,y), so that the slope through (2,9) and that point is equal to 6x. so,
(y-9)/(x-2) = 6x
y-9 = 6x(x-2)
3x² - 9 = 6x² - 12x
0 = 3x² - 12x + 9
x² - 4x + 3 = 0
(x-3)(x-1) = 0
x = 3, y = 3(3)² = 27
x = 1, y = 3(1)² = 3
so (3,27) and (1,3)