If the graph has a point of inflection at (2,0)..?

NetherCraft 0

If the graph of y = x³ + ax² + bx – 8 has a point of inflection at (2,0), what is the value of b?

(A) 0

(B) 4

(C) 8

(D) 12

(E) The value of b cannot be determined from the given information.

I put E because I figured b doesn’t last until the second derivative, so it is impossible to find the value. I know the 2nd derivative = 0 at 2, but am I able to use that information for this problem? Any help would be appreciated.

4 Answers

  • If y(x) = x³ + a.x² + b.x – 8 then

    dy/dx = 3.x² + 2.a.x + b = 0

    d2y/dx2 = 6.x + 2.a = 0

    for a stationary point which is also a point of inflexion. If this is at x = 2, the second equation gives a = -6. Put this in the first equation (for dy/dx) to get

    3.x² – 12.x + b = 0 = 3.(x – 2)²

    so b = 12: note that points of inflexion are produced where two or more roots are identical. In this case I think all three are at x = 2.

  • As you know, you can find the 2nd derivative, which is 6x+2a. Since you know it equals zero at x=2, find a.

    6(2)+2a=0

    2a=-12

    a=-6

    Then, plug “a” back into the original equation:

    y=x^3-6x^2+bx-8

    Since the point (2,0) exists, you can use this fact to solve for b. Plug in 2 for x and 0 for y in the original.

    f(2)= 8-24+2b-8=0

    24=2b

    b=12

    And that is your answer. D.

  • y” = 6x+2a = 0

    So when x = 2, we have a = -6

    So equation becomes x^3 -6x^2 +bx -8 = 0

    But f(2) = 0, so 8 – 24+2x -8 = 0 –> b = 12

  • Take the derivative. Then take the derivative returned. The x value the place this 2d derivative (f”(x) or y”) equation equals 0 factors the x value of your factor of inflection. (The y value of the inflection factor could be got here upon using the unique equation.)

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