From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	To determine all the points on the curve <math>y=3x^{2}-12x+9</math> which also lie on the ''x''-axis we substitute the equation of the ''x''-axis i.e. <math>y=0</math> in the equation of the curve and obtain that ''x'' must satisfy
-	<center> [[Bild:2_3_9c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>0 = 3x^{2}-12x+9\,\textrm{.}</math>}}
-	[[Bild:2_3_9_c.gif|center]]	+
		+	After dividing by 3 and completing the square the right-hand side is
		+
		+	{{Displayed math||<math>x^{2}-4x+3 = (x-2)^{2} - 2^{2} + 3 = (x-2)^{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 ''x''-axis are (1,0) and (3,0).
		+
		+
		+	[[Image:2_3_9_c.gif|center]]

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Current revision

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).