Posted by Jake on .
An equation to the graph of y=x^3+3x^2+2 at its point of inflection is:
Please help. I am in urgent need.
Math (Calculus AB) -
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
Which of these two lines (with slope = -3) gives y=4 at x=-1?