Solution 2.3:9a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (13:14, 29 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
A point lies on the
+
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
-
<math>x</math>
+
-
-axis if it has
+
-
<math>y</math>
+
-
-coordinate
+
-
<math>0</math>
+
-
and we therefore look for all the points on the curve
+
-
<math>y=x^{\text{2}}-\text{1}</math>
+
-
where
+
-
<math>y=0</math>, i.e. all points which satisfy the equation
+
 +
{{Displayed math||<math>0=x^{2}-1\,\textrm{.}</math>}}
-
<math>0=x^{\text{2}}-\text{1}</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 \text{1}</math>, which means that the points of intersection are
+
-
<math>\left( -1 \right.,\left. 0 \right)</math>
+
-
and
+
-
<math>\left( 1 \right.,\left. 0 \right)</math>.
+
[[Image:2_3_9_a.gif|center]]
[[Image:2_3_9_a.gif|center]]

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).


Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.3:9a"