Brain Teaser 7: Line of Symmetry, Maxima (or, Minima) Of Quadratic Function in Terms Of a, b and c

The Brain Teaser: For quadratic function y = ax² + bx + c, show that ax² + bx + c ≡ a(x – s)² + m, where x = s = - (b/2a) is its line of symmetry; m = c - (b²/4a) is its maxima (or minima, if a > 0)

Any quadratic function graph has a line of symmetry x = s and its maxima (or, minima whichever is applicable) lies at the point of intersection of the graph and its line of symmetry.

In other words, if you substitute x = s into y = ax² + bx + c, you would get the value of maxima or minima i.e. m. Hence, if you know the value of the line of symmetry of a quadratic function, you can use it to calculate its maxima or minima in lieu of completing the square method or the dy/dx method. 

But what is the value of s in terms of a, b and c? And, what is m in terms of a, b and c? This, in essence, is what this Brain Teaser 7 is all about!

In a week or so, I will post the working in getting ax² + bx + c ≡ a(x – s)² + m, where:

1)       x = s = - (b/2a)  (Do you not see this line of symmetry inside the Quadratic Root Formula?)

2)      m = c - (b²/4a) 

Test these with y = (x + 2)(x – 3) = x² – x – 6, where a = 1, b = -1 and c = -6 and

a    *   The line of symmetry x = s = - (b/2a) = - (-1/2) = ½; and

*   Minima (a = 1 > 0), m = c - (b²/4a) = -6 – (1/4x1) = -6 – ¼ = - 6.25

Have fun with maths!

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

Simple proof:

If the quadratic function y = ax² + bx + c is expressed as

                                       y = a(x – s)² + m,

show that:

·        its line of symmetry is x = s = -b/2a

·        its minima (if a > 0) or maxima (if a < 0)

m = c – (b²/4a) = y when x = s = -b/2a

Solution:

            y = ax² + bx + c

               = a[x² + (b/a)x] + c

               = a[(x + b/2a)² – (b/2a)²] + c

               = a(x + b/2a)² - a(b/2a)² + c

               = a(x + b/2a)² + c – b²/4a

Thus, if y = a(x – s)² + m,

              s = -b/2a, and

              m = c – (b²/4a)

Point of maxima or minima occurs when x = s (i.e. at its line of symmetry).

At x = s = -b/2a, maximum or minimum value of y is as follows:

     y =  ax² + bx + c

        = a(-b/2a)² + b(-b/2a) + c

        = b²/4a – b²/2a + c

        = b²/4a – 2b²/4a + c

        = c – (2b²/4a - b²/4a)

        = c – (b²/4a) = m,   as in the expression y = a(x – s)² + m         …(Shown)