Solution 2.3:9c
From Förberedande kurs i matematik 1
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)
