Question 1125140
Use the vertex form of the quadratic,
{{{y=a(x-h)^2+k}}}
{{{y=a(x-2)^2+5}}}
Solve for {{{a}}} using the second point,
{{{-3=a(0-2)^2+5}}}
{{{-3=a(4)+5}}}
{{{4a=-8}}}
{{{a=-2}}}
So,
{{{y=-2(x-2)^2+5}}}
Expanding,
{{{y=-2(x^2-4x+4)+5}}}
{{{y=-2x^2+8x-8+5}}}
{{{y=-2x^2+8x-3}}}
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*[illustration 231.PNG].