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Given the graph the student enters the  formula

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{eqnline(a,b,x2,y2,v)}

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Give the equation of the quadratic graph shown above.

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We know that the graph crosses the \$x\$-axis at both \$(\\var{a},0)\$ and \$(\\var{b},0)\$. Since this is a quadratic, we know our equations has two roots, and by the previous observation, they are at \$\\var{a}\$ and \$\\var{b}\$. Hence we can write our equation as \$\\simplify{y=(x-{a})(x-{b})}\$ which simplifies to \$\\simplify{y=x^2-({a}+{b})x+({a}*{b})}\$.

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To find the coefficients of the turning point of the quadratic, we know the x-coordinate of the turning point will correspond to the solution to \$dy/dx=0\$. So we get \$\\simplify{2x-({a}+{b})}=0\$ hence \$\\simplify{x=({a}+{b})/2}\$. We substitute this value of x back into the equation of the quadratic to find the corresponding y-coordinate.

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Write the equation of the graph.

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\$g(x)=\\;\$[[0]]

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