// Numbas version: exam_results_page_options {"name": "Graph1", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"maxRuns": 100, "condition": ""}, "parts": [{"gaps": [{"answer": "{a}x+{b}", "variableReplacements": [], "marks": 1, "showCorrectAnswer": true, "checkingtype": "absdiff", "checkvariablenames": false, "showpreview": true, "answersimplification": "all", "expectedvariablenames": [], "scripts": {}, "type": "jme", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "vsetrangepoints": 5, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst"}], "type": "gapfill", "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "showFeedbackIcon": true, "prompt": "

Write the equation of the line in the diagram. The line described by your equation will also be drawn on the diagram.

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$y=\\;$[[0]]

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

\n

The above bar graph shows wheat output in kilotons each year.

\n

Which year had the highest wheat output?

", "rulesets": {}, "variables": {"y2": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "y2", "definition": "x2*a+b"}, "a": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "a", "definition": "random(-4..4 except 0)"}, "b": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "b", "definition": "random(-6..6 except [0,a])"}, "x2": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "x2", "definition": "random(-3..3 except -1..1)"}}, "metadata": {"description": "

There are copious comments in the definition of the function eqnline about the voodoo needed to have a JSXGraph diagram interact with the input box for a part.

", "licence": "Creative Commons Attribution 4.0 International"}, "extensions": ["jsxgraph"], "preamble": {"js": "", "css": ""}, "advice": "\n

First Method.

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You are given that the line goes through $(0,\\var{b})$ and $(-1,\\var{b-a})$ and the equation of the line is of the form $y=ax+b$

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Hence:

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1) At $x=0$ we have $y=\\var{b}$, and this gives $\\var{b}=a \\times 0 +b =b$ on putting $x=0$ into $y=ax+b$.

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So $b=\\var{b}$.

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2) At $x=-1$ we have $y=\\var{b-a}$, and this gives $\\var{b-a}=a \\times (-1) +b =\\simplify[all,!collectNumbers]{-a+{b}}$ on putting $x=-1$ into $y=ax+b$.

\n

On rearranging we obtain $a=\\simplify[all,!collectNumbers]{{b}-{b-a}}=\\var{a}$.

\n

So $a=\\var{a}$.

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So the equation of the line is $\\simplify{y={a}*x+{b}}$.

\n

Second Method.

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The equation $y=ax+b$ tells us that the graph crosses the $y$-axis (when $x=0$) at $y=b$.

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So looking at the graph we immediately see that $b=\\var{b}$.

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$a$ is the gradient of the line and is given by the change from $(-1,\\var{b-a})$ to $(0,\\var{b})$:

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\$a=\\frac{\\text{Change in y}}{\\text{Change in x}}=\\frac{\\simplify[all,!collectNumbers]{({b-a}-{b})}}{-1-0}=\\var{a}\$

\n\n", "variable_groups": [], "type": "question", "contributors": [{"name": "Raymond Corrigan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2229/"}]}]}], "contributors": [{"name": "Raymond Corrigan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2229/"}]}