Math (Calculus AB)
posted by Jake on .
An equation to the graph of y=x^3+3x^2+2 at its point of inflection is:
a) y=3x+1
b)y=3x7
c)y=x+5
d)y=3x+1
e)y=3x+7
Please help. I am in urgent need.

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) = 36=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?