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The function $f: y=\\simplify{{a}*x^2+{b}*x+c}$ has $\\var{yv}$ as an extreme value.
", "advice": "A quadratic function $f:y=ax^2+bx+c$ has a parabolic graph and reaches it extremal value at the vertex.
\nThe $x$-value of the vertex is found as $x=-\\frac{b}{2a}=\\simplify{{-b}/{2*a}}=\\var{xv}$.
\nThe extremal value itself is given by $f(\\var{xv})=\\var{yv-c}+c$. Since this must be equal to $\\var{yv}$, we find $\\var{yv-c}+c=\\var{yv}$, hence $c=\\var{c}$.
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", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "yv+a*xv^2", "description": "constante term van kwadratische functie
", "templateType": "anything", "can_override": false}, "xv": {"name": "xv", "group": "Ungrouped variables", "definition": "random(-5 .. 5#1)", "description": "x-coordinate of the vertex
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\n$c=$[[0]]
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