// Numbas version: exam_results_page_options {"name": "Francis '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": [{"functions": {"eqnline": {"definition": "// This function creates the board and sets it up, then returns an\n// HTML div tag containing the board.\n \n// The line is described by the equation \n// y = a*x + b\n\n// This function takes as its parameters the coefficients a and b,\n// and the coordinates (x2,y2) of a point on the line.\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('400px','400px',\n{boundingBox: [-13,22,13,-22],\n axis: false,\n showNavigation: false,\n grid: true\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,2],{\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,2],{\ndrawLabels: true,\nlabel: {offset: [-20, 0]},\nminorTicks: 0\n});\n\n// mark the two given points - one on the y-axis, and one at (x2,y2)\n\n\n\n\nboard.create('functiongraph',[function(x){ return (x-a)*(x-b);},-13,13]);\n\nreturn div;", "type": "html", "language": "javascript", "parameters": [["a", "number"], ["b", "number"], ["x2", "number"], ["y2", "number"]]}}, "ungrouped_variables": ["a", "x2", "b", "y2", "c"], "name": "Francis 's copy of Quadratic graph - student finds equation", "tags": ["graphs", "jsxgraph", "Jsxgraph"], "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 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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{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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