Post a New Question

Math (Calculus AB)

posted by .

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) -

    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?

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question