// Numbas version: exam_results_page_options {"name": "Quadratic graph vertical stretch 1", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Quadratic graph vertical stretch 1", "tags": [], "metadata": {"description": "

Given th original formula the student enters the transformed formula

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Ready to use

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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The Blue graph shows a graph of a quadratic equation, $f(x)=x^2$

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The Blue graph has been streched vertically onto the red graph $g(x)$, type in the new function definition:

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The Blue graph has been streched vertically onto the black graph $h(x)$, type in the new function definition:

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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 red line.

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

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Write the equation of the line black line.

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

\n

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