Question 1194535
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The two zeros are given as x=2 and x=6.<br>
You could start with that, knowing that the equation is of the form<br>
{{{f(x)=a(x-2)(x-6)}}}<br>
Then determine the value of a using any of the given points.  e.g.,<br>
{{{f(0)=a(-2)(-6)=12a=3}}} --> {{{a=1/4}}}<br>
Then one form of the equation is<br>
{{{f(x)=(1/4)(x-2)(x-6)}}}<br>
Reformat it into any form you want.<br>
Alternatively, with the zeros at x=2 and x=6, you know the axis of symmetry is x=4; and since f(4)=-1, you know the vertex is (4,-1).  Then you could start with vertex form,<br>
{{{f(x)=a(x-h)^2+k}}}
{{{f(x)=a(x-4)^2-1}}}<br>
Again determine the value of a by using any of the given data points (except (4,-1))<br>
{{{f(0)=a(-4)^2-1 = 16a-1 = 3}}}
{{{16a=4}}}
{{{a=1/4}}}<br>
And so<br>
{{{f(x)=(1/4)(x-4)^2-1}}}<br>
Again reformat that in any way you need to.<br>