1. Complete The Square

2. #1 Make sure your equation is in standard form:
Y = aX2+bX+c or 0 = ax2+bx+c
Y = 2X2 + 4X + 8
0 = 2X2 + 4X + 8
#2 Set Y = 0
(2)
#3 Make the coefficient of X2 equal to 1 by factoring, NOT division. (Don’t forget about this factor later)
0 = X2 + 2X + 4
#4 Move the “C” value to make way for a value that makes a perfect square.
-4 = X2 + 2X

3. #4 Move the “C” value to make way for a value that makes a perfect square.
-4 = X2 + 2X
#5 Find the new value of “C” by the following method:
-3 = X2 + 2X +1
#6 We now have a perfect square on the right, so change it to the factored form.
Hint: It’s just
-3 = (X+1)2

4. #6 We now have a perfect square on the left, so change it to the factored form.
-3 = (X+1)2
#7 Now set the equation back to zero.
0 = (X+1)2 +3
(2)
#8 Did you factor in the beginning? Better distribute it back.
0 = 2(X+1)2 +6
Y = 2(X+1)2 +6
#9 We set Y = 0, so 0 = Y
We did it, we completed the square and converted the equation into Vertex Form!

5. So …. What’s so great about that?

6. Our Vertex Form Equation Y = 2(X + 1)2 + 6
The Format of Vertex Form
Y = a(X - h)2 +k
Since our “a” is positive, the graph is going up!
(h,k) is the vertex! So our vertex is:
(-1,6)
Set Y = 0 and solve for X to find the Roots! (Where the graph intercepts the X-Axis).
Note: “a” is not the same “a” value of the standard form.

7. Finding the Roots
Given
Set Y =0
Subtract 6 from Both Sides
Divide Both Sides by 2
Take the Square Root of Both Sides
Subtract 1 from Both Sides
The Roots are
&

8. Since you can not take an even root of a negative number, the roots are imaginary. This means our graph never goes through the x-axis.