Question

Find the points on the curve y=x^3-4x^2+1 where the tangent is parallel to y+4x=0.

y'=3x^2-8x

y+4x=3x^2-8x
y=3x^2-12x
y=3x(x-4)

Ya... I need some help

The answers are (2, -7) (2/3, -67/27)

Answer 1

y = (-4) x, slope is -4
so
where is dy/dx = -4
dy/dx = 3 x^2 - 8x
so where is
3 x^2 - 8 x + 4 = 0 ???
solve quadratic
hummm, not hard because we know 3*2 + 2 = 8
(3x-2)(x-2) = 0
x = 2/3 or x = 2
go back and get y values from original cubic curve

Answer 2

hey
y+4x=0 is a straight line
y = -4x + 0 in form y = m x + b
-4 is the slope !!!!

Answer 3

hmmm. Didn't think of solving for x?
y+4x=0 means y = -4x. So,
3x^2-8x = -4x
3x^2-4x=0
x=0 or 4/3

But that is not the problem. You don't care where the two graphs intersect. You care where the slopes are equal. so, you really want
3x^2-8x = -4
3x^2-8x+4 = 0
(3x-2)(x-2)=0
So, the is -4 at (2/3,-13/27) and (2,-7)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+x%5E3-4x%5E2%2B1+,+y%3D-4(x-2%2F3)-13%2F27,+y%3D-4(x-2)-7