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)
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
hey
y+4x=0 is a straight line
y = -4x + 0 in form y = m x + b
-4 is the slope !!!!
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
To find the points on the curve where the tangent is parallel to the given line y + 4x = 0, we need to find the values of x that satisfy two conditions: the derivative of the curve at that point should exist, and it should be equal to the slope of the given line (-4 in this case).
Let's start by finding the derivative of the curve y = x^3 - 4x^2 + 1. Taking the derivative, we get:
y' = 3x^2 - 8x
Now, we need to find the values of x for which y' = -4 (the slope of the given line).
Setting y' equal to -4, we have:
3x^2 - 8x = -4
Rearranging the equation, we get:
3x^2 - 8x + 4 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 3, b = -8, and c = 4. Plugging in these values, we get:
x = (-(-8) ± √((-8)^2 - 4 * 3 * 4)) / (2 * 3)
x = (8 ± √(64 - 48)) / 6
x = (8 ± √16) / 6
x = (8 ± 4) / 6
x = (12 or 4) / 6
x = 2 or 2/3
So the values for x that satisfy y' = -4 are x = 2 and x = 2/3.
Now we need to find the corresponding y-values. We can plug these x-values back into the original equation y = x^3 - 4x^2 + 1 to find the corresponding y-values.
For x = 2:
y = (2)^3 - 4(2)^2 + 1
y = 8 - 16 + 1
y = -7
So we have one point (2, -7) on the curve where the tangent is parallel to y + 4x = 0.
For x = 2/3:
y = (2/3)^3 - 4(2/3)^2 + 1
y = 8/27 - 16/9 + 1
y = -67/27
So we have another point (2/3, -67/27) on the curve where the tangent is parallel to y + 4x = 0.
Therefore, the points on the curve y = x^3 - 4x^2 + 1 where the tangent is parallel to y + 4x = 0 are (2, -7) and (2/3, -67/27).