Algebra : Quadratic Functions : Maxima and minima values

An expression, f(x) or y = ax² + bx + c is a quadratic function. The shape is a parabola and depends on whether a is positive or negative value. If a > 0, the curve has a minimum value at A and if a<0, the curve has a maximum value at B. IMG_20150518_082328

For example, y = x² – 4x + 3. Since a = +1, this curve will have a minimum value. Using completing the square method :    y = x² – 4x + 5 + (-4/2)² – (-4/2)² = (x – 2)² + 5 – (-2)² = (x – 2)² + 1.

As (x – 2)² ≥ 0 for all x, minimum value of y is 1, and minimum value occurs when (x – 2)² = 0, or x = 2. Point (2, 1) is the minimum point of the curve. Also when x = 0, y = 3, so the curve cuts the y-axis at (0, 3). The sketch of the curve showing the minimum point is :

IMG_20150518_091017

Try this question :

1) Express y = 10 – 2x – x² in the form y = a – (x + b)² . Hence, find the maximum value of y and the value of x at this maximum value. Also sketch the curve.

Answer :

1) y = 10 -2x – x² = – (x² + 2x – 10) = – (x² + 2x + (2/2)² – 10 – (2/2)² = – 〈(x + 1)² – 11〉

        = 11 – (x + 1)²

    Since (x + 1)² ≥ 0 for all values of x

    ∴ − (x + 1)² ≤ 0 and thus, 11 − (x + 1)² ≤ 11

     ∴ max. value of y is 11 and the max. point is (−1, 11).

     When x = 0, y = 10.

IMG_20150520_082616

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