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Find the $x$ and $y$ values that satisfy both of the following equations. That is, find the points of intersection of the two curves.
\n$y$ | \n$=$ | \n$\\simplify{{grad}x+{yint}}$ | \n$(1)$ | \n
$y$ | \n$=$ | \n$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$ | \n$(2)$ | \n
$x_1=$ [[0]], $y_1=$ [[1]] and $x_2=$ [[2]], $y_2=$ [[3]]
\nNote: To input your answer please ensure that $x_1<x_2$.
\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": true, "variableReplacements": [], "maxValue": "root1", "minValue": "root1", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": true, "variableReplacements": [], "maxValue": "ansy1", "minValue": "ansy1", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": true, "variableReplacements": [], "maxValue": "root2", "minValue": "root2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": true, "variableReplacements": [], "maxValue": "ansy2", "minValue": "ansy2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "steps": [{"prompt": "There are many ways to solve these equations simultaneously. Here is one method.
\n$y$ | \n$=$ | \n$\\simplify{{grad}x+{yint}}$ | \n$(1)$ | \n
$y$ | \n$=$ | \n$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$ | \n$(2)$ | \n
Substitute the expression for $y$ given in $(1)$ into $(2)$:
\\[\\simplify{{grad}x+{yint} =x^2+{quadxcoeff}x+{quadccoeff}}\\]
Since we have a quadratic here we get everything onto one side:
\\[0=\\simplify{x^2+{sroots}x+{proots}}\\]
There are various ways to solve a quadratic, in this particular case we can factorise the quadratic:
\n\\[(\\simplify{x-{root1}})(\\simplify{x-{root2}})=0\\]
\nTherefore, $x=\\var{root1},\\,\\var{root2}$.
\n
Now for $x=\\var{root1}$, we can determine the corresponding $y$ value by substituting $x=\\var{root1}$ into either equation $(1)$ or $(2)$. Below we substitute into $(1)$:
$y$ | \n$=$ | \n$\\simplify[!collectnumbers]{{grad}({root1})+{yint}}$ | \n
\n | $=$ | \n$\\var{ansy1}$ | \n
Now for $x=\\var{root2}$, so we can determine the corresponding $y$ value by substituting $x=\\var{root2}$ into either equation $(1)$ or $(2)$. Below we substitute into $(1)$:
\n$y$ | \n$=$ | \n$\\simplify[!collectnumbers]{{grad}({root2})+{yint}}$ | \n
\n | $=$ | \n$\\var{ansy2}$ | \n
Therefore the values that satisfy equations $(1)$ and $(2)$ are $x_1=\\var{root1}$, $y_1=\\var{ansy1}$ and $x_2=\\var{root2}$, $y_2=\\var{ansy2}$.
\nIn other words, the two curves intersect at the points $(\\var{root1},\\var{ansy1})$ and $(\\var{root2},\\var{ansy2})$.
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