Lösung 2.3:9a
Aus Online Mathematik Brückenkurs 1
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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 | 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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>0=x^{2}-1\,\textrm{.}</math>}} |
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>. | 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>. | ||
Version vom 08:34, 22. Okt. 2008
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).
