find the equation of the tangent to y= 1-3x+12x^2-8x^3
parrarel to tangent (1,2)
2007-09-30 09:35:38 · 3 answers · asked by edik 4 in Science & Mathematics ➔ Mathematics
i keep gettin y=(-3x+5)
but the answer at the back says it y=(-3x+1)
2007-09-30 09:51:28 · update #1
y' = - 24x^2 + 24x - 3
y'(1) = -24 + 24 - 3 = -3 = m
2 = (-3)(1) + b
b = 5
y = -3x + 5
If you were asking for the line parallel to the tangent line of y at x=1, going through the point (1,2).
You didn't really ask the question well, so I'm assuming that's what you were asking.
2007-09-30 09:43:21 · answer #1 · answered by Will 3 · 0⤊ 0⤋
y = 1 - 3x + 12x^2 - 8x^3
the slope of the tangent = y'(1)
y'(x) = -3 +24x - 24x^2
y'(1) = -3 +24 - 24
= -3
The tangent parallel will have same slope
the equation of tangent = y-y1=m(x-x1)
=>y - 2 = -3(x - 1)
=> y - 2 = -3x + 3
=> y + 3x = 5
2007-09-30 16:50:56 · answer #2 · answered by mohanrao d 7 · 0⤊ 1⤋
I agree with you.
2007-09-30 17:12:51 · answer #3 · answered by Tony 7 · 0⤊ 1⤋