// Numbas version: exam_results_page_options {"name": "Helge's copy of Adrian's copy of Quadratic graph - student finds equation", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "tags": [], "statement": "

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

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The above graph shows a graph of a quadratic equation, it is your task to find this equation.

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You are given the two points of the curve with the x axis, \$(\\var{b},0)\$ and \$(\\var{a},0)\$, and the \$y\$-intercept at \$(0,\\var{c})\$ as indicated on the diagram.

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Students enter equation and turning point

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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 line in the diagram.

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\$y=\\;\$[[0]]

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Find the coordinates of the turning point of this quadratic

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\$x=\$[[0]]

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\$y=\$[[1]]

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