Answer
Function on Left: $y=2(x-3)^2$
Function on Right: $y=\frac{-1}{2}x^2-2x+1$
Work Step by Step
Left Function:
Using the basic factored form of the parabola $y=a(x-b)^2+c$, we can determine the equation of the graph from the two points given (vertex and other point). The vertex is (3,0), which tells us that the graph has shifted 3 units to the right of the origin and 0 units up - hence b = 3 and c = 0. So we get $y=a(x-3)^2$ . To find a, just sub (4,2) into the equation to get $2=a(4-3)^2 $ -> $a=2$. Therefore $y=2(x-3)^2$
Right Function:
Use the other form of the parabola: $y=ax^2+bx+c$. From the graph, we can tell the $y-int = 1$, hence $c=1$, so the equation becomes $y=ax^2+bx+1$. Now sub both the points (-2,2) and (1,-2.5) into the equation to get a pair of simultaneous equations:
(1) $2 = 4a -2b$
(2) $-2.5=a+b$ --> $a=-b-2.5$
sub (2) into (1)
$2=4(-b-2.5)-2b$
$2=-4b-10-2b$
$-6b = 12$
$b=-2$ ... sub into (1)
$4a = -2$
$a = \frac{-1}2$
And finally the equation is: $y=\frac{-1}{2}x^2-2x+1$
