Find the equation of the tangent line to the given function at the given point. Show all work.
2006-10-03 13:51:04 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x) = x^4 - 3x^2 + 2
f`(x) = 4x ^ 3 - 6x
f'(1) = 4 - 6
f' (1) = -2
therefore,
y-0 = -2 ( x - 1)
y = -2(x-1)
=)
2006-10-03 13:57:33 · answer #1 · answered by sam t 1 · 2⤊ 0⤋
Equation of the tangent line is a the 1st derivative. Not sure how to actually get the equation, it has been a few years.
y' = 4x^3-6x is the first derivative. Now I know you have to plug the point in there somehow, but I forget exactly how. Perhaps someone else could finish?
2006-10-03 14:00:42 · answer #2 · answered by abbubaca 1 · 0⤊ 0⤋
y=x^4-3x^2+2
y = 1- 9 + 2
y = 1-9+2
y=-6
2006-10-03 14:01:06 · answer #3 · answered by pbuilder_novis 3 · 0⤊ 0⤋
y' = 4x^3 - 6x, slope at (1,0) is -2
the equation y/(x-1) = -2. y = -2x+2.
2006-10-03 15:03:33 · answer #4 · answered by shamu 2 · 0⤊ 0⤋
y'=4x^3-6x
y'(1,0)=-2
y= -2x+b
0=-2+b
y=-2x+2
2006-10-03 14:01:16 · answer #5 · answered by rwbblb46 4 · 1⤊ 0⤋