Forming Perfect Squares

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Presentation transcript:

Using the Algebraic Method to Find Optimum Values of Quadratic Functions

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.

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

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.

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

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.

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 – 12 + 2 = (x2 + 2x + 1) – 1 + 2 = (x + 1)2 + 1 ∵ Coefficient of x2 = 1 > 0 ∴ The minimum value of y = x2 + 2x + 2 is 1.

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.