Question 884252
{{{y=ax^2 + bx + c}}}
When {{{x=0}}},{{{y=-3}}}
{{{-3=a(0)^2+b(0)+c}}}
{{{c=-3}}}
So now you already know {{{c}}}
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When {{{x=1}}},{{{y=0}}}
{{{0=a(1)^2+b(1)-3}}}
1.{{{a+b=3}}}
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.
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When {{{x=2}}},{{{y=5}}}
{{{5=a(2)^2+b(2)-3}}}
{{{4a+2b=8}}}
2.{{{2a+b=4}}}
Now you have two equations in two unknowns and you should be able to solve.
Subtract eq. 1 from eq. 2,
{{{2a+b-a-b=4-3}}}
{{{a=1}}}
Now back substitute to find {{{b}}}
{{{1+b=3}}}
{{{b=2}}}
So now you have all three values.
{{{highlight_green(y=x^2+2x-3)}}}