Solution 2.3:9c
From Förberedande kurs i matematik 1
m  (Lösning 2.3:9c moved to Solution 2.3:9c: Robot: moved page)  | 
				|||
| Line 1: | Line 1: | ||
| - | {  | + | To determine all the points on the curve   | 
| - | <  | + | <math>y=3x^{2}-12x+9</math>  | 
| - | {{  | + | which also lie on the   | 
| + | <math>x</math>  | ||
| + | -axis we substitute the equation of the   | ||
| + | <math>x</math>  | ||
| + | -axis i.e.   | ||
| + | <math>y=0</math>  | ||
| + | in the equation of the curve and obtain that   | ||
| + | <math>x</math>  | ||
| + | must satisfy   | ||
| + | |||
| + | |||
| + | <math>3x^{2}-12x+9=0</math>  | ||
| + | |||
| + | |||
| + | After dividing by   | ||
| + | <math>3</math>  | ||
| + | and completing the square the right-hand side is   | ||
| + | |||
| + | |||
| + | <math>x^{2}-4x+3=\left( x-2 \right)^{2}-2^{2}+3=\left( x-2 \right)^{2}-1</math>  | ||
| + | |||
| + | |||
| + | and thus the equation has solutions  | ||
| + | |||
| + | |||
| + | <math>x=2\pm 1,</math>  | ||
| + | i.e.   | ||
| + | <math>x=2-1=1</math>  | ||
| + | and   | ||
| + | <math>x=2+1=3.</math>  | ||
| + | |||
| + | |||
| + | The points where the curve cut the   | ||
| + | <math>x</math>  | ||
| + | -axis are  | ||
| + | |||
| + | |||
| + | <math>\left( 1 \right.,\left. 0 \right)</math>  | ||
| + | and   | ||
| + | <math>\left( 3 \right.,\left. 0 \right)</math>  | ||
| + | |||
| + | |||
| + | |||
[[Image:2_3_9_c.gif|center]]  | [[Image:2_3_9_c.gif|center]]  | ||
Revision as of 12:11, 21 September 2008
To determine all the points on the curve \displaystyle y=3x^{2}-12x+9 which also lie on the \displaystyle x -axis we substitute the equation of the \displaystyle x -axis i.e. \displaystyle y=0 in the equation of the curve and obtain that \displaystyle x must satisfy
\displaystyle 3x^{2}-12x+9=0
After dividing by 
\displaystyle 3
and completing the square the right-hand side is 
\displaystyle x^{2}-4x+3=\left( x-2 \right)^{2}-2^{2}+3=\left( x-2 \right)^{2}-1
and thus the equation has solutions
\displaystyle x=2\pm 1,
i.e. 
\displaystyle x=2-1=1
and 
\displaystyle x=2+1=3.
The points where the curve cut the 
\displaystyle x
-axis are
\displaystyle \left( 1 \right.,\left. 0 \right)
and 
\displaystyle \left( 3 \right.,\left. 0 \right)

