Completing the square?

Dec 2009
8
0
How do I convert an equation from standard form to vertex form?

Standard: y= x^2+6x +5

Vertex form: ?
 
May 2010
16
6
Sydney - Australia
hello SpaceGhost,

The vertex form of a quadratic is given by y = a(x – h)2 + k, where (h, k) is the vertex.

To convert from general to vertex you basically take the 'loose term' out of the general form and do simple 'Completing the Square" :

\(\displaystyle y=x^2+6x+5\)
\(\displaystyle y-5 = x^2+6x\)
\(\displaystyle y-5+(\frac {6}{2})^2=x^2+6x+(\frac{6}{2})^2\)

\(\displaystyle \frac {y-20+36}{4}=(x+3)^2\)
\(\displaystyle y=4(x+3)^2-16\)

that should be the vertex form :)
 
Dec 2009
8
0
Gah I still don't get it. Why did you get y-5 in the second line and 6/2? I still just dpn't get it. ):
 
May 2010
16
6
Sydney - Australia
From the first line i moved the +5 over to the other side and it became -5. To perform the 'Completing the Square' Method, there must be the squared term and the linear term (which is in our case \(\displaystyle 6x\)). The 2nd step is to add to both sides of the equation (so equality is established) the square of the co-efficient of the linear term and divided by 2 (in our case, the co-efficient of the linear term was 6, so we divide it by two and then square the whole thing).
Step 3, you simplify the term inside the brackets now if you can. (in our case 6/2 = 3
Step 4: Now we have a perfect square: (\(\displaystyle x^2+6x+9\)
Step 5: factorise it, you don't really need to because you have already done it before in the previous step. To factorise it, you just take out that halved co-efficient and your 'x', add them together or subtract from another, depending on the sign of the linear term (here we had a positive +6x) so we add the the two terms, then we square because its a perfect square: \(\displaystyle (x+3)^2\)

then you simplify