From Förberedande kurs i matematik 1

Revision as of 09:05, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:9a moved to Solution 2.3:9a: Robot: moved page) ← Previous diff		Current revision (13:14, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	A point lies on the ''x''-axis if it has ''y''-coordinate 0 and we therefore look for all the points on the curve <math>y=x^{2}-1</math> where <math>y=0</math>, i.e. all points which satisfy the equation
-	<center> [[Image:2_3_9a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>0=x^{2}-1\,\textrm{.}</math>}}
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		+	This equation has solutions <math>x=\pm 1</math>, which means that the points of intersection are <math>(-1,0)</math> and <math>(1,0)</math>.
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	[[Image:2_3_9_a.gif|center]]		[[Image:2_3_9_a.gif|center]]

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Current revision

A point lies on the x-axis if it has y-coordinate 0 and we therefore look for all the points on the curve \displaystyle y=x^{2}-1 where \displaystyle y=0, i.e. all points which satisfy the equation

\displaystyle 0=x^{2}-1\,\textrm{.}

This equation has solutions \displaystyle x=\pm 1, which means that the points of intersection are \displaystyle (-1,0) and \displaystyle (1,0).