Question 459899
<pre>y = -2x² + 8x + 3

Factor the coefficient of the x² term out of the x-terms on the right
[Do not factor out the x, but only the numerical coefficient]

y = -2(x² + 4x) + 3

Multiply the coefficient of x by 1/2, then square what you get.

4(1/2) = 2, then 2² = 4

Add and then subtract that number inside the parentheses:
[This just amounts to adding 0 which does not change the 
value]

y = -2(x² + 4x + 4 - 4) + 3  

Change the parentheses to brackets so you can insert parentheses
around the first three terms:

y = -2[(x² + 4x + 4) - 4] + 3

Factor the trinomial in the parentheses which should turn
out to be the square of a binomial:

y = -2[(x + 2)(x + 2) - 4] + 3

y = -2[(x + 2)² - 4] + 3

Remove the brackets by multiplying the -2 by each of
the terms inside the bracket, keeping the (x + 2)² intact:

y = -2(x + 2)² + 8 + 3

y = -2(x + 2)² + 11

That's it.  

Compare to 

y = a(x - h)² + k

and you will see that the vertex is (-2,11)

-------------------------------------------

y = 3x² - 6x + 5

Factor the coefficient of the x² term out of the x-terms on the right
[Do not factor out the x, but only the numerical coefficient]

y = 3(x² - 2x) + 5

Multiply the coefficient of x by 1/2, then square what you get.

-2(1/2) = -1, then (-1)² = 1

Add and then subtract that number inside the parentheses:
[This just amounts to adding 0 which does not change the 
value]

y = 3(x² - 2x + 1 - 1) + 5  

Change the parentheses to brackets so you can insert parentheses
around the first three terms:

y = 3[(x² - 2x + 1) - 1] + 5

Factor the trinomial in the parentheses which should turn
out to be the square of a binomial:

y = 3[(x - 1)(x - 1) - 1] + 5

y = 3[(x - 1)² - 1] + 5

Remove the brackets by multiplying the 3 by each of
the terms inside the bracket, keeping the (x + 2)² intact:

y = 3(x - 1)² - 3 + 5

y = 3(x - 1)² + 2

That's it.  

Compare to 

y = a(x - h)² + k

and you will see that the vertex is (1,2)

Edwin</pre>