Answer
See explanation.
Work Step by Step
(a) First derivative is $2bx +3 a x^{2}+c=f^{\prime}(x)$
Second derivative is $2 b+6 a x=f^{\prime \prime}(x)$
The inflection point is the root of the second derivative, which is $-\frac{b}{3 a}=x$
(b) Assume that $x_{1}, x_{2}$ and $x_{3}$ are the zeros ( $x$ -intercepts) of the function. And then $f(x)=a\left(-x_{1}+x\right)\left(-x_{2}+x\right)\left(-x_{3}+x\right)$
The first derivative is then (using the chain rule): $f^{\prime}(x)=a\left(-x_{1}+x\right)(-$ $\left.x_{2}+x\right)+a\left(-x_{1}+x\right)\left(-x_{3}+x\right)+a\left(-x_{2}+x\right)\left(-x_{3}+x\right)$
Second derivative is then (using the chain rule): $f^{\prime \prime}(x)=\left(-x_{1}+x\right)2 a+$ $\left(-x_{2}+x\right)2 a+\left(-x_{3}+x\right)2 a=-2 a\left(x_{1}+x_{2}+x_{3}\right)+6 a x$
The inflection point is the root of the second derivative, which is $x=$ $\frac{2 a\left(x_{3}+x_{1}+x_{2}\right)}{6 a}=\frac{x_{3}+x_{2}+x_{1}}{3}$, and this means that the inflection point is the average of the three $x$ -intercepts. This is (1,0).
(c) Determine the zeros of $f(x)=x\left(-3 x+2+x^{2}\right)=(x-2)(x-1)x$, which are $x=2, x=1$ and $x=0 .$ The average is of the three zeros is then $\frac{0+1+2}{3}=1$ so the inflection point is at $1=x$
$f^{\prime \prime}(x)=-6+6 x$ which is positive if $x>1$ and negative if $x<1 .$ So the function $f$ is concave up on $(1,+\infty)$ and concave down on $(-\infty, 1)$
