We want to get y = a(x-h)²+k
y = x²-3x+6 in the form
Put brackets around the first two terms on the
right,
y = [x²-3x]+6
and since the coefficient of x² is 1 put 1
in front of the parentheses. Normally you
would factor out the coefficient of x²:
y = 1[x²-3x]+6
We complete the square inside the brackets:
1. Multiply the coefficient of x, which is -3
by , getting .
2. Square getting =
3. Add , which is really 0, inside the
brackets:
y = 1[x²-3x+-]+6
Factor the first three terms inside the bracket like this
x²-3x+ = (x-)(x-) = (x-)²
y = 1[(x-)²-]+6
Remove the bracket
y = 1(x-)²-+6
Write the 6 as
y = 1(x-)²-+
y = 1(x-)²+
y = a(x-h)²+k
The vertex is (h,k) = (,)
Edwin