1. Using the Algebraic Method to Find Optimum Values of Quadratic Functions

2. Forming Perfect Squares
Consider x2 + 6x.
By adding , we have
x2 + 6x
= x2 + 2(3)x + 32
◄ x2 + 2mx + m2 ≡ (x + m)2
= (x + 3)2
a perfect square
This process is called completing the square.

3. Completing the square To complete the squares for x2 + kx and x2 – kx,
add to each expression. Then

4. How to find the optimum value of the quadratic function y = ax2 + bx + c?
Convert y = ax2 + bx + c to the form y = a(x – h)2 + k by completing the square first.

5. Optimum values of quadratic functions
y = ax2 + bx + c
y = a(x – h)2 + k
completing
the square b 2a +  2

6. Optimum values of quadratic functions
y = ax2 + bx + c
y = a(x – h)2 + k
completing
the square
If a > 0, then the minimum value of y is k when x = h.
If a < 0, then the maximum value of y is k when x = h.

7. Can you find the optimum value of the function y = x2 + 2x + 2?
Complete the square for x2 + 2x. 2 ø ö ç è æ
Add , i.e. 12.
= x2 + 2x + 12 –
= (x2 + 2x + 1) – 1 + 2
= (x + 1)2 + 1
∵ Coefficient of x2 = 1 > 0
∴ The minimum value of y = x2 + 2x + 2 is 1.

8. Follow-up question
Find the optimum value of the quadratic function y = –4x2 + 24x – 15, and the corresponding value of x.
y = –4x2 + 24x – 15
= –4(x2 – 6x) – 15
Complete the square for x2 – 6x. 2 6 ø ö ç è æ
Add , i.e. 32.
= –4(x2 – 6x + 32 – 32) – 15
= –4(x2 – 6x + 9) + 36 – 15
= –4(x – 3)2 + 21
∵ Coefficient of x2 = –4 < 0
∴ The maximum value of y = –4x2 + 24x – 15 is 21 when x = 3.