Solution 2.3:9a
From Förberedande kurs i matematik 1
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| - | {{  | + | A point lies on the   | 
| - | <  | + | <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   | ||
| + | |||
| + | |||
| + | <math>0=x^{\text{2}}-\text{1}</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]]  | ||
Revision as of 11:43, 21 September 2008
A point lies on the \displaystyle x -axis if it has \displaystyle y -coordinate \displaystyle 0 and we therefore look for all the points on the curve \displaystyle y=x^{\text{2}}-\text{1} where \displaystyle y=0, i.e. all points which satisfy the equation
\displaystyle 0=x^{\text{2}}-\text{1}
This equation has solutions
\displaystyle x=\pm \text{1}, which means that the points of intersection are 
\displaystyle \left( -1 \right.,\left. 0 \right)
and
\displaystyle \left( 1 \right.,\left. 0 \right).

