Find the equation of the tangent line to the curve?

Question

y=1-6 x-6 x^2
at the point (1,-11).

y= x/ (3-x)
at point (0,0)

methods would help drastically

Answer 1

y = 1 - 6x - 6x^2

find the derivative
y' = -6 - 12x

plug 1 in to find the slope of the tangent line
y' = -6 - 12(1)
y' = -18

y = mx + b
-11 = -18(1) + b
b = 7

y = -18x + 7 <== answer


use quotient rule:
d/dx (u/v) = (u'v - v'u) / v^2

y' = [ d/dx(x) (3 - x) - d/dx (3 - x) (x) ] / (3 - x)^2
y' = [ 3 - x - (-1)(x) ] / (3 - x)^2
y' = ( 3 - x + x) / (3 - x)^2
y' = 3 / (3 - x)^2

plug 0 in
y' = 3/(3 - 0)^2
y' = 3/(3)^2
y' = 1/3

y = mx + b
0 = 1/3(0) + b
b = 0

y = x/3 <== answer

Answer 2

y = 1 - 6x - 6 x^2 . . . . differentiating
y ' = -6 - 12 x . . . . . substitute x = 1
y ' = - 6 - 12 = - 18
slope of line = - 18
- 18 = (y + 11) / ( x - 1)
- 18 (x - 1) = (y + 11)
- 18 x + 18 = y + 11
18 x + y - 7 = 0

y = x / ( 3 - x)
y ' = [ (3 - x) - x (-1) ] /( 3 - x)^2 . . . . .. differentiating
y ' = 3 / 9 = 1/3 . . . . . . substituting x = 0
slope = 1/3
1/3 = y /x
x = 3y
x - 3y = 0