Question 344386
Both zeros must be equidistant from the line of symmetry.
If {{{x=3}}} is a zero, so is {{{x=-3}}}
{{{f(x)=a(x-3)(x+3)}}}
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Use the point ({{{0}}},{{{6}}}) to find {{{a}}}.
{{{f(0)=a(0-3)(0+3)=6}}}
{{{-9a=6}}}
{{{highlight(a=-2/3)}}}
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{{{f(x)=-(2/3)(x-3)(x+3)}}}
{{{highlight_green(f(x)=-(2/3)(x^2-9))}}}
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{{{drawing(300,300,-10,10,-10,10,grid(1),circle(-3,0,0.3),circle(3,0,0.3),circle(0,6,0.3),graph(300,300,-10,10,-10,10,-(2/3)*(x^2-9)))}}}