Solution 2.3:9b

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The points of intersection are those points on the curve which also lie on the
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The points of intersection are those points on the curve which also lie on the ''x''-axis, i.e. they are those points which satisfy both the equation of the curve
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<math>~x</math>
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<math>y=x^{2}-5x+6</math> and the equation of the ''x''-axis <math>y=0</math>,
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-axis, i.e. they are those points which satisfy both the equation of the curve
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<math>y=x^{\text{2}}-\text{5}x+\text{6}</math>
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and the equation of the
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<math>~x</math>
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-axis
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<math>y=0</math>,
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{{Displayed math||<math>\left\{\begin{align}
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y&=x^{2}-5x+6\,,\\
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y&=0\,\textrm{.}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{matrix}
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This system of equations gives directly that <math>y=0</math> and that <math>x</math> must satisfy the quadratic equation <math>x^{2}-5x+6=0\,</math>. By completing the square, we obtain that the left-hand side is
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y=x^{\text{2}}-\text{5}x+\text{6} \\
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y=0\quad \quad \quad \quad \\
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\end{matrix} \right.</math>
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{{Displayed math||<math>\begin{align}
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x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt]
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&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt]
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&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{1}{4}
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\end{align}</math>}}
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This system of equations gives directly that
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and this gives that the equation has solutions <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, i.e. <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> and <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.
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<math>y=0</math>
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and that
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<math>~x</math>
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must satisfy the second-order equation
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<math>x^{\text{2}}-\text{5}x+\text{6}=0</math>
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. By completing the square, we obtain that the left-hand side is
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The intersection points are therefore (2,0) and (3,0).
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<math>\begin{align}
 
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& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
 
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& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
 
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\end{align}</math>
 
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<center>[[Image:2_3_9b-2(2).gif]]</center>
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and this gives that the equation has solutions
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<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.
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<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>
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and
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<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.
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The intersection points are therefore
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<math>\left( 2 \right.,\left. 0 \right)</math>
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and
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<math>\left( 3 \right.,\left. 0 \right)</math>.
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{{NAVCONTENT_START}}
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<center> [[Image:2_3_9b-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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Current revision

The points of intersection are those points on the curve which also lie on the x-axis, i.e. they are those points which satisfy both the equation of the curve \displaystyle y=x^{2}-5x+6 and the equation of the x-axis \displaystyle y=0,

\displaystyle \left\{\begin{align}

y&=x^{2}-5x+6\,,\\ y&=0\,\textrm{.} \end{align}\right.

This system of equations gives directly that \displaystyle y=0 and that \displaystyle x must satisfy the quadratic equation \displaystyle x^{2}-5x+6=0\,. By completing the square, we obtain that the left-hand side is

\displaystyle \begin{align}

x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{1}{4} \end{align}

and this gives that the equation has solutions \displaystyle x=\tfrac{5}{2}\pm\tfrac{1}{2}, i.e. \displaystyle x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2 and \displaystyle x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3.

The intersection points are therefore (2,0) and (3,0).


Image:2_3_9b-2(2).gif
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