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^36x^2+bx8
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)= 824+2b8=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.)