Question 356349
Since it has zeros at {{{x=2}}} and {{{x=3}}}, the quadratic equation has the form,
{{{f(x)=a(x-2)(x-3)}}}
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Use the last point ({{{0}}},{{{3}}}) to find a.
{{{f(0)=a(0-2)(0-3)=3}}}
{{{a(6)=3}}}
{{{a=1/2}}}
{{{f(x)=(1/2)(x-2)(x-3)}}}
{{{f(x)=(1/2)(x^2-3x-2x+6)}}}
{{{f(x)=(1/2)(x^2-5x+6)}}}
{{{highlight(f(x)=(1/2)x^2-(5/2)x+3)}}}
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{{{drawing(300,300,-3,5,-3,5,circle(3,0,0.2),circle(2,0,0.2),circle(0,3,0.2),grid(1),graph(300,300,-3,5,-3,5,(1/2)(x^2-5x+6)))}}}