User:Volvagia356/Sandbox: Difference between revisions

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Revision as of 13:22, 24 June 2010

$f(x)=ax^{2}+bx+c\,$

$f(x)=ax^{2}+bx+c$

$f(x)=a(x-p)^{2}+q$

$x+p=0$

$x=-p$

${\frac {b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

$4=a(-2)^{2}+b(-2)+4.5$

$4=4a-2b+4.5$

$4=4a+4.5$

$4a=-0.5$

$a={\frac {-1}{8}}$

$y=-{\frac {1}{8}}x^{2}+4.5$

${\frac {1}{2}}$

${\frac {1}{2}}=0a^{2}+0b+c$

$c={\frac {1}{2}}$

$y=-{\frac {1}{8}}x^{2}+0.5$

${\frac {-1}{8}}(x-2)^{2}+{\frac {1}{2}}$

$\int _{-2}^{2}-{\frac {1}{8}}x^{2}+0.5\,dx$

@@ Line 1: / Line 1: @@
+==Description of Parabola==
-Maths Project Testing
-<math>\int_{-2}^{2}-\frac{1}{8}x^2+0.5\, dx</math>
 <math>f(x)=ax^2+bx+c\,</math>
@@ Line 14: / Line 11: @@
 <math>\frac{b\pm\sqrt{b^2-4ac}}{2a}</math>
+==First Function==
 <math>4=a(-2)^2+b(-2)+4.5</math>
@@ Line 25: / Line 22: @@
 <math>a=\frac{-1}{8}</math>
-<math>y=\frac{1}{8}x^2+4.5</math>
+<math>y=-\frac{1}{8}x^2+4.5</math>
+==Second Function==
 <math>\frac{1}{2}</math>
@@ Line 33: / Line 30: @@
 <math>c=\frac{1}{2}</math>
-<math>y=\frac{1}{8}x^2+0.5</math>
+<math>y=-\frac{1}{8}x^2+0.5</math>
+==Third Function==
 <math>\frac{-1}{8}(x-2)^2+\frac{1}{2}</math>
+==Finding the Surface Area==
+<math>\int_{-2}^{2}-\frac{1}{8}x^2+0.5\, dx</math>