The curve passes through the point (-1, 0). This means that this point satisfies the equation of the curve.
0 = a(-1^3) + 2(-1^2) + a^2(-1) + b
0 = -a + 2 + a^2 + b
a^2 - a + b = -2
The curve has a minimum at (-1, 0). This means that the derivative of the curve at x = -1 is 0
dy/dx = 3ax^2 + 4x + a^2
At x = -1,
3a - 4 + a^2 = 0
a^2 + 3a - 4 = 0
(a + 4)(a-1) = 0
a = -4 or a = 1
When a = -4
(-4)^2 - (-4) + b = -2
16 + 4 + b = -2
b = -18
When a = 1
1^2 - 1 + b = -2
1 - 1 + b = -2
b = -2
In general, when a question asks to solve for two variables, you need to set up two equations and then solve the system of equations to determine the values of those variable. By extension, if a question asks to find the values of n unknowns, you need to set up a system of n equations and solve that system.
Michael W.
Ah, tricky tricky. For a point to be a local minimum, the derivative of the function has to switch from negative to positive (First Derivative Test, indicating that the curve changes from decreasing to increasing), orrrrr the second derivative of the function has to be positive (Second Derivative Test, indicating that the curve is concave up). Just by setting the first derivative to zero, we don't know which one of those is happening, so we need to go one step further.
I'm sure there's a way to use the First Derivative Test, but I feel like the Second Derivative Test will work better here:
y'' = 6ax + 4
When x is -1, we plug that in, and we get y'' = -6a + 4. When a is -4, we get a positive second derivative...so that one works. When a is 1, we get a negative second derivative, so that'd indicate a local maximum, not a local minimum, so that one's no good.
Does that help?
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08/15/16
Daniel K.
08/15/16