A parabolic graph is given. The question is to determine the equation of the graph. Non-calculator. Advice is given.
"}, "preamble": {"css": "", "js": ""}, "advice": "The graph crosses the $x$-axis in two places, namely at $(\\var{r1},0)$ and $(\\var{r2},0)$. Therefore, our quadratic can be written as $a(x-\\var{r1})(x-\\var{r2})$, but we still need to determine $a$. To find $a$, you need to pick a value of $x$ (which is not $\\var{r1}$ or $\\var{r2}$) and use the graph to determine what $f(x)$ is. Then you plug in these numbers into the equation and solve for $a$. Unfortunately, there is no fixed choice for $x$ so feedback on this exact question is not available. However, we present an example from a different question:
\n\n
Imagine after the first step you reached $f(x) = a(x-3)(x+2)$. By looking at the graph we saw that $f(4) = 12$. Plugging $x=4$ into our equation we get:
$12 = f(4) = a(4-3)(4+2) = a \\cdot 1 \\cdot 6 = 6a$, so $a = 2$.
\nThen the final answer in this example would be $f(x) = 2(x-3)(x+2)$. (Depending on the wording of the question, you may have to expand the brackets too. In this example, expanding the brackets gives $f(x) = 2x^2 - 2x - 12$.)
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", "parts": [{"customName": "", "type": "gapfill", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "marks": 0, "showFeedbackIcon": true, "gaps": [{"customName": "", "type": "jme", "vsetRangePoints": 5, "answer": "{a}(x-{r1})(x-{r2})", "showFeedbackIcon": true, "failureRate": 1, "checkingType": "absdiff", "checkVariableNames": false, "showCorrectAnswer": true, "useCustomName": false, "customMarkingAlgorithm": "", "answerSimplification": "all", "variableReplacementStrategy": "originalfirst", "marks": "4", "valuegenerators": [{"name": "x", "value": ""}], "checkingAccuracy": 0.001, "scripts": {}, "extendBaseMarkingAlgorithm": true, "showPreview": true, "unitTests": [], "variableReplacements": [], "vsetRange": [0, 1]}], "sortAnswers": false, "prompt": "{plotgraph(r1,r2,a)}
\nWrite the equation of the quadratic graph.
\n$y(x)=\\;$[[0]]
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