Lösung 2.3:9c
Aus Online Mathematik Brückenkurs 1
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Version vom 14:16, 22. Okt. 2008
To determine all the points on the curve \displaystyle y=3x^{2}-12x+9 which also lie on the x-axis we substitute the equation of the x-axis i.e. \displaystyle y=0 in the equation of the curve and obtain that x must satisfy
| \displaystyle 0 = 3x^{2}-12x+9\,\textrm{.} |
After dividing by 3 and completing the square the right-hand side is
| \displaystyle x^{2}-4x+3 = (x-2)^{2} - 2^{2} + 3 = (x-2)^{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 x-axis are (1,0) and (3,0).
