From Förberedande kurs i matematik 1

Revision as of 09:05, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:9b moved to Solution 2.3:9b: Robot: moved page) ← Previous diff		Revision as of 11:57, 21 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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-	{{NAVCONTENT_START}}	+	The points of intersection are those points on the curve which also lie on the
-	<center> [[Image:2_3_9b-1(2).gif]] </center>	+	<math>~x</math>
-	{{NAVCONTENT_STOP}}	+	-axis, i.e. they are those points which satisfy both the equation of the curve
		+	<math>y=x^{\text{2}}-\text{5}x+\text{6}</math>
		+	and the equation of the
		+	<math>~x</math>
		+	-axis
		+	<math>y=0</math>,
		+
		+
		+	<math>\left\{ \begin{matrix}
		+	y=x^{\text{2}}-\text{5}x+\text{6} \\
		+	y=0\quad \quad \quad \quad \\
		+	\end{matrix} \right.</math>
		+
		+
		+	This system of equations gives directly that
		+	<math>y=0</math>
		+	and that
		+	<math>~x</math>
		+	must satisfy the second-order equation
		+	<math>x^{\text{2}}-\text{5}x+\text{6}=0</math>
		+	. By completing the square, we obtain that the left-hand side is
		+
		+
		+	<math>\begin{align}
		+	& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
		+	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
		+	\end{align}</math>
		+
		+
		+	and this gives that the equation has solutions
		+
		+	<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.
		+	<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>
		+	and
		+	<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.
		+
		+	The intersection points are therefore
		+	<math>\left( 2 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 3 \right.,\left. 0 \right)</math>.
		+
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:2_3_9b-2(2).gif]] </center>		<center> [[Image:2_3_9b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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Revision as of 11:57, 21 September 2008

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

\displaystyle \left\{ \begin{matrix} y=x^{\text{2}}-\text{5}x+\text{6} \\ y=0\quad \quad \quad \quad \\ \end{matrix} \right.

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

\displaystyle \begin{align} & x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\ & =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\ \end{align}

and this gives that the equation has solutions

\displaystyle x=\frac{5}{2}\pm \frac{1}{2}, i.e. \displaystyle x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2 and \displaystyle x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3.

The intersection points are therefore \displaystyle \left( 2 \right.,\left. 0 \right) and \displaystyle \left( 3 \right.,\left. 0 \right).