Posted by **Jake** on Monday, January 24, 2011 at 11:13pm.

An equation to the graph of y=x^3+3x^2+2 at its point of inflection is:

a) y=-3x+1

b)y=-3x-7

c)y=x+5

d)y=3x+1

e)y=3x+7

Please help. I am in urgent need.

- Math (Calculus AB) -
**MathMate**, Tuesday, January 25, 2011 at 12:27am
I suppose the question is:

"A *line tangent* to the graph of y=x^3+3x^2+2 at its point of inflection is"

First find the derivatives:

y' = 3x²+6x

y" = 6x+6

At the point of inflexion, y"=0, or x=-1.

Then the slope at x=-1 is

y'(-1) = 3-6=-3

There are only two choices of lines that have a slope of -3.

From these, we calculate

y(-1) = (-1)³+3(-1)²+2

=-1+3+2

=4

Which of these two lines (with slope = -3) gives y=4 at x=-1?

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