a) a=0 makes f(x) a constant function
b) \(\text{the "top" of } f(x) \text{ is }f(0)=0 \\\text{ the "top" of }g(x) \text{ is } f\left(\dfrac b 6\right) =-\dfrac{b^2}{12} \)
c) \(a(-3)^2 + 8 = 5\\ 9a = -3\\ a=-\dfrac 1 3\)
d) \(\text{the coefficient of the }x^2 \text{ term controls the narrowness of the parabola}\\ \text{the larger absolute value of the coefficient the narrower the parabola}\\ \text{a negative value produces a downward facing parabola, thus}\\ a < -3 \text{ will produce a downward facing parabola that is narrower than g(x)}\)
e) \(3(-2)^2 - b(-2) = 0\\ 12 +2 b = 0\\ b=-6\)
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