Answer
Left graph: $y=2x^{2}-12x+18$
Right graph: $y=-x^{2}-2.5x+1$
Work Step by Step
Left graph:
$y=a(x-b)^{2}+c$
vertex$(3,0)$ $b=3, c=0$
at $(4,2), 2=a(4-3)^{2}$
$a=2$
$\therefore y=2(x-3)^{2}=2x^{2}-12x+18$
Right graph:
$y=ax^{2}+bx+c$
at $(0,1), 1=0^{2}a+(0)b+c,$
$c=1$
at $(-2,2), 2=(-2)^{2}a-2b+1,$
$4a-2b=1$ (i)
at $(1, -2.5), -2.5=1^{2}a+(1)b+1$
$a+b=-3.5$ (ii)
(ii)$\times2$: $2a+2b=-7$ (iii)
(i)$+$(iii): $6a=-6, a=-1 \rightarrow$ii)
$-1+b=-3.5, b=-2.5$
$\therefore y=-x^{2}-2.5x+1$
