Skillnad mellan versioner av "2.2 Lösning 2b"
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Taifun (Diskussion | bidrag) m (Created page with "<math> y = -3\,x^2 + 2\,x - 12 </math> i intervallet <math> -2 \leq x \,\leq\, 2 </math>") |
Taifun (Diskussion | bidrag) m |
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| + | <math> f(2) = -3\cdot 2^2 + 2\cdot 2 - 12 = -3\cdot 4 + 4 - 12 = -12 + 4 - 12 </math> | ||
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| + | <math> f(-2) = - [-3\cdot (-2)^2 + 2\cdot (-2) - 12] = </math> | ||
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| + | <math> \Delta y = </math> | ||
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| + | <math> = - [-3\cdot 4 - 4 - 12] = </math> | ||
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| + | <math> {\Delta y \over \Delta x} = {f(3) \, - \, f(2) \over 2 - (-2)} = {5\cdot 3 + 23 \, - \, (5\cdot 2 + 23) \over 1} = </math> | ||
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| + | <math> = 15+23 \, - \, (10+23) = 38-33 = 5 </math> | ||
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<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> | ||
Versionen från 1 maj 2011 kl. 15.40
\( f(2) = -3\cdot 2^2 + 2\cdot 2 - 12 = -3\cdot 4 + 4 - 12 = -12 + 4 - 12 \)
\( f(-2) = - [-3\cdot (-2)^2 + 2\cdot (-2) - 12] = \)
\( \Delta y = \)
\( = - [-3\cdot 4 - 4 - 12] = \)
\( {\Delta y \over \Delta x} = {f(3) \, - \, f(2) \over 2 - (-2)} = {5\cdot 3 + 23 \, - \, (5\cdot 2 + 23) \over 1} = \)
\( = 15+23 \, - \, (10+23) = 38-33 = 5 \)
\( y = -3\,x^2 + 2\,x - 12 \) i intervallet \( -2 \leq x \,\leq\, 2 \)
