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how to form the quadratic function into y=ax^2+bx+c

 Jul 22, 2014

Best Answer 

 #1
avatar+26400 
+6

how to form the quadratic function into y=ax^2+bx+c  ?

$$\begin{array}{rcl}
y&=&
ax^2+bx+c\\\\
y&=&
a\left(
x^2+{b\over a}x\right)
+c\\\\
y&=&
a\left(
x^2+{b\over a}x \textcolor[rgb]{1,0,0}{
+ ({b\over 2a})^2 - ({b\over 2a})^2 }
\right)
+c\\\\
y&=&
a\left[(x+{b\over 2a})^2 - ({b\over 2a})^2
\right]
+c\\\\
y&=&a (x+{b\over 2a})^2 - {b^2\over 4a}+c\\\\
y&=&a (x+{b\over 2a})^2 + c - {b^2\over 4a}
\end{array}$$

 Jul 22, 2014
 #1
avatar+26400 
+6
Best Answer

how to form the quadratic function into y=ax^2+bx+c  ?

$$\begin{array}{rcl}
y&=&
ax^2+bx+c\\\\
y&=&
a\left(
x^2+{b\over a}x\right)
+c\\\\
y&=&
a\left(
x^2+{b\over a}x \textcolor[rgb]{1,0,0}{
+ ({b\over 2a})^2 - ({b\over 2a})^2 }
\right)
+c\\\\
y&=&
a\left[(x+{b\over 2a})^2 - ({b\over 2a})^2
\right]
+c\\\\
y&=&a (x+{b\over 2a})^2 - {b^2\over 4a}+c\\\\
y&=&a (x+{b\over 2a})^2 + c - {b^2\over 4a}
\end{array}$$

heureka Jul 22, 2014
 #2
avatar+118723 
0

Morgan Tud contributed this thread a while back.

http://web2.0calc.com/questions/magick-of-the-quadratic#rr2

 Jul 22, 2014

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