2.2 Lösning 2b - Versionshistorik

Taifun den 7 augusti 2014 kl. 14.53

2014-08-07T14:53:08Z

2014-08-07T14:52:01Z

2014-08-07T14:50:25Z

2011-05-01T16:00:07Z

2011-05-01T15:58:13Z

2011-05-01T15:54:15Z

2011-05-01T15:40:32Z

2011-05-01T15:20:43Z

Created page with "<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>"

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<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>

← Äldre version		Versionen från 7 augusti 2014 kl. 14.53
Rad 4:		Rad 4:


−	<big><big><math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} </math></big></big> <math> = \, 2 </math>	+	<big><big><math> {\Delta y \over \Delta x} = {f(2)\,-\,f(-2) \over 2\,-\,(-2)} = {-20\,-\,(-28) \over 4} = {8 \over 4} </math></big></big> <math> = \, 2 </math>

← Äldre version		Versionen från 7 augusti 2014 kl. 14.52
Rad 4:		Rad 4:


−	<big><big><math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} ~~= 2~~ </math></big></big>	+	<big><big><math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} </math></big></big> <math> = \, 2 </math>

← Äldre version		Versionen från 7 augusti 2014 kl. 14.50
Rad 3:		Rad 3:
	<math> f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -28 </math>		<math> f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -28 </math>

−	<math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} = 2 </math>	+
		+	<big><big><math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} = 2 </math></big></big>

@@ Rad 1: / Rad 1: @@
 <math> f(2) = -3\cdot 2^2 + 2\cdot 2 - 12 = -3\cdot 4 + 4 - 12 = -12 + 4 - 12 = -20 </math>
-<math> f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -12 - 4 - 12 = -28 </math>
+<math> f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -28 </math>
 <math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} = 2 </math>

@@ Rad 3: / Rad 3: @@
 <math> f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -12 - 4 - 12 = -28 </math>
-<math> \Delta y = </math>
+<math> {\Delta y \over \Delta x} = {f(2) - f(-2) \over 2 - (-2)} = {-20 - (-28) \over 4} = {8 \over 4} = 2 </math>
-−
-<math> {\Delta y \over \Delta x} = {f(3) \, - \, f(2) \over 2 - (-2)} = {5\cdot 3 + 23 \, - \, (5\cdot 2 + 23) \over 1} = </math>
-−
-−
-<math> = 15+23 \, - \, (10+23)  = 38-33 = 5 </math>
-−
-<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>

← Äldre version		Versionen från 1 maj 2011 kl. 15.40
Rad 1:		Rad 1:
		+	<math> f(2) = -3\cdot 2^2 + 2\cdot 2 - 12 = -3\cdot 4 + 4 - 12 = -12 + 4 - 12 </math>
		+
		+	<math> f(-2) = - [-3\cdot (-2)^2 + 2\cdot (-2) - 12] = </math>
		+
		+	<math> \Delta y = </math>
		+
		+	<math> = - [-3\cdot 4 - 4 - 12] = </math>
		+
		+	<math> {\Delta y \over \Delta x} = {f(3) \, - \, f(2) \over 2 - (-2)} = {5\cdot 3 + 23 \, - \, (5\cdot 2 + 23) \over 1} = </math>
		+
		+
		+	<math> = 15+23 \, - \, (10+23) = 38-33 = 5 </math>
		+
	<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>		<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>