// Numbas version: exam_results_page_options {"name": "Graph transforms TEST 1", "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('600px','600px',\n{boundingBox: [-12,12,12,-12],\n axis: false,\n showNavigation: false,\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,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);},-12,12]);\nboard.create('functiongraph',[function(x){ return 2*(x-a)*(x-b);},-12,12],{ strokeColor: 'red'});\n\nreturn div;", "type": "html", "language": "javascript", "parameters": [["a", "number"], ["b", "number"], ["x2", "number"], ["y2", "number"]]}}, "ungrouped_variables": ["a", "x2", "b", "y2", "c"], "name": "Graph transforms TEST 1", "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 line in the diagram.

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$g(x)=\\;$[[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 Blue graph shows a graph of a quadratic equation, $f(x)=x^2-\\var{y2}x+\\var{c}$

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

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e.g   $g(x)=Af(x\\pm B)$

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Need to randomise the transformation (currently 3) perhaps between -3 to +3

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Also axis should be variable so as to see whole of graph.

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Given the original formula the student enters the transformed formula

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