Posted by mary on Tuesday, March 23, 2010 at 11:31pm.
let's first find the equation of the tangent at (-2,8)
(you should have checked that (-2,8) actually lies on f(x) )
f'(x) = 3x^2 + 6x - 1
at the point given
f'(-2) = 12 - 12 - 1 = -1
so y = -x + b, but (-2,8) lies on it
8 = 2 + b
b = 6
the tangent is y = -x + 6
let's now intersect that with f(x)
x^3 + 3x^2 - x + 2 = - x + 6
x^3 + 3x^2 - 4 = 0
we know that (x+2) is a factor since x = -2 was a solution
by synthetic division
x^3 + 3x^2 - 4 = (x+2)(x^2 + x - 2)
x^3 + 3x^2 - 4 = (x+2)(x+2)(x-1)
so we have a double root at x = -2 (this always happens at the tangent contact point)
and x = 1
when x = 1, y = 5
so Q must be (1,5)
b) I don't quite understand the question, but perhaps you can take it from here
Find the coordinates of all points on yhe graph of y= 1-x^2 at which the tangent line passes through the point (2,0).
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