gurmies said:
Answer is a = 4 and b = 4
or a = 2 b = -4
x^3 + 4x^2 + 4x
d/dx (x^3 + 4x^2 + 4x ) = 3x^2 + 8x + 4
Now, for a stationary point, derivative must be equal to 0.
3x^2 + 8x + 4 = 0
(3x+2)(x+2) = 0
x = -2/3 or/ x = -2 (which is what we need)
Similar sort of method for a = 2, b = -4
If that question came up in a test, that method would be extremely ineffective and time wasting
The fastest way to do it would be
f(x) = x^3 + ax^2 + bx
f'(x) = 3x^2 + 2ax + b
Since stationary point at x = -2
0 = 3(-2)^2 + 2a(-2) + b
0 = 12 - 4a + b
b = 4a - 12 ____(1)
Also, since (-2, 0) is a point of the curve
0 = -8 + 4a - 2b
4a = 2b + 8 ____(2)
Sub (1) -> (2)
4a = 2(4a - 12) + 8
4a = 8a - 24 + 8
-4a = -16
a = 4
Sub into (1)
b = 4(4) - 12
b = 4
Therefore a = 4 and b = 4 are the only ones that work
This method is much better/faster than the trial and error method shown above
Also, less chance of making a mistake (ie. shown above as well)
The mistake in the above post is not clear on first sight, as it seems to fulfill the condition that x = -2. However, the question states that the stationary point is at (-2, 0), not when x = -2
SO if a = 2 and b = -4, then the y value of when x = -2 would be 8, which would make the stationary point (-2, 8). This is incorrect but it's quite hard to see that using gurmie's method. It's safest just to use my method
