Partially Preposterous Parabolic Perplexing Problem Part II

We know that the Vertex form of a parabola is $y=a(x-h)^2+k$ if the vertex is $h,k$ .

Plugging in $(4,2)$ as the vertex, we get the equation is $y=a(x-4)^2+2$ .

Plugging in point $(2,0)$ in for x and y, we get that $0=a(2-4)^2+2$ , or $a=-\dfrac{1}{2}$ .

Now we expand $y=-\dfrac{1}{2}(x-4)^2+2$ to get $y=-\dfrac{1}{2}x^2+4x-6$ .

Our answer is therefore $-\dfrac{1}{2}\cdot 4\cdot -6=\boxed{12}$ and we are done.

Using vertex form $y = a(x-4)^2+2 0=a(2-4)^2+2$ solve then $a=-1/2$ Plug this to the original vertex form $y=-.5(x-4)^2+2$ $Y=-.5x^2 +4x-6$ So now just multiply values of abc $-.5*4*-6=12$

We know that the parabola $y=ax^{2}+bx+c$ passes through $(4,2)$ and $(2,0)$ , so we have:

$2=a(4)^{2}+b(4)+c$

$2=16a+4b+c$

And:

$0=a(2)^{2}+b(2)+c$

$0=4a+2b+c$

Rest the second equation from the first one:

$2=12a+2b$

Also, we know that the vertex is $(4,2)$ , hence:

$-\frac{b}{2a}=4$

$b=-8a$

Substitute in the previous equation:

$2=12a+2(-8a)$

$2=12a-16a$

$2=-4a$

$\boxed{a=-\frac{1}{2}}$

Solve for $b$ :

$b=-8(-\frac{1}{2}$

$\boxed{b=4}$

And finally, solve for $c$ by substituting on the first equation the values of $a$ and $b$ :

$2=16(-\frac{1}{2})+4(4)+c$

$2=-8+16+c$

$2=8+c$

$\boxed{c=-6}$

We found that the parabola is: $y=-\frac{1}{2}x+4x-6$ , so $abc=(-\frac{1}{2})(4)(-6)=\boxed{12}$

Since parabola is symmetric about its vertex, if it passes through the point (2,0) then it has to pass through (6,0) as well...substituting these two points and the vertex in given equation will give three simple linear equations in terms of a,b and c..solving them we can get values of a, b ,c and hence abc....