Calculus: Early Transcendentals 8th Edition

Left graph: $y=2x^{2}-12x+18$ Right graph: $y=-x^{2}-2.5x+1$
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) $-1+b=-3.5, b=-2.5$ $\therefore y=-x^{2}-2.5x+1$