// Numbas version: exam_results_page_options {"name": "Quadratic graph equation", "extensions": ["geogebra", "jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {"eqnline": {"definition": "// This function creates the board and sets it up, then returns an\n// HTML div tag containing the board.\n\n//Put in your values of x here\n\nvar x_min = -10;\nvar x_max = 10;\nvar y_min = -12;\n//var y_min = -(a*b+2);\n//var y_min = (((a*b)-((a+b)/2)^2)-2);\n\nvar y_max = 10;\n\n// First, make the JSXGraph board.\n// The function provided by the JSXGraph extension wraps the board up in \n// a div tag so that it's easier to embed in the page.\nvar div = Numbas.extensions.jsxgraph.makeBoard('600px','600px',\n//{boundingBox: [-8,10,8,-10],\n {boundingBox: [x_min,y_max,x_max,y_min], \n axis: false,\n showNavigation: true,\n grid: true\n});\n\n\n\n\n// div.board is the object created by JSXGraph, which you use to \n// manipulate elements\nvar board = div.board; \n\n// create the x-axis.\nvar xaxis = board.create('line',[[0,0],[1,0]], { strokeColor: 'black', fixed: true});\nvar xticks = board.create('ticks',[xaxis,1],{\n drawLabels: true,\n label: {offset: [-4, -10]},\n minorTicks: 0\n});\n\n// create the y-axis\nvar yaxis = board.create('line',[[0,0],[0,1]], { strokeColor: 'black', fixed: true });\nvar yticks = board.create('ticks',[yaxis,1],{\ndrawLabels: true,\nlabel: {offset: [-20, 0]},\nminorTicks: 0\n});\n\n\n // PUT YOUR FUNCTION HERE\n\n\n\nboard.create('functiongraph',[function(x){ return (x+a)*(x+b);},x_min,x_max]);\n\n\n\nreturn div;", "type": "html", "language": "javascript", "parameters": [["a", "number"], ["b", "number"], ["x2", "number"], ["y2", "number"], ["v", "number"]]}}, "ungrouped_variables": ["a", "x2", "b", "y2", "c", "v"], "name": "Quadratic graph equation", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

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 quadratic graph.

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

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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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Given the original formula the student enters the transformed formula

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