# calculus

posted by on .

let f be the function
f(x) = x^3 + 3x^2 - x + 2

a. the tangent to the graph of f at the point P = (-2,8) intersects the graph of f again at the point Q. Find the coordinates of point Q.

b. Find the coordinates of point R, the inflection point of the graph of f

c. Show that the segment QR divides the region between the graph of f and its tangent at P into two regions whose areas are in the ratio of 16/11

• calculus - ,

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

• calculus - ,

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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