Please help if you can:
Find all the values of x for which the tangent like to curve
y= (2x^3)-x^2 is parrallel to a line with the slope of 4.
2006-10-25 14:59:45 · 2 answers · asked by Beef 5 in Science & Mathematics ➔ Mathematics
Take the derivative to get the slope, then solve for values = 4.
y = 2x^3 - x^2
y' = 6x^2 - 2x
6x^2 - 2x = 4
6x^2 - 2x - 4 = 0
3x^2 - x - 2 = 0
Factor:
(x-1)(3x + 2) = 0
x = 1
x = -2/3
2006-10-25 15:07:20 · answer #1 · answered by Puzzling 7 · 5⤊ 0⤋
y =2x^3 - x^2
Finding the tangent to a curve at a point is the same as saying "find the slope of the curve at that point". To do that, take the derivative.
y' = 6x^2 - 2x
or, slope = 6x^2 - 2x.
All lines parallel to a line with slope 4, also have slope 4.
So, 4 = 6x^2 - 2x
Rearranging gives :
6x^2 - 2x - 4 = 0
or,
3x^2 - x - 2 = 0
or,
(3x + 2)(x - 1) = 0
Therefore, all the values of x are -2/3 and 1.
2006-10-25 22:26:01 · answer #2 · answered by falzoon 7 · 3⤊ 0⤋