Find the domain and the range of the quadratic function whos

Find the domain and the range of the quadratic function whose maximum value is 15 at x = -14. The domain of f is: _____________ The range of f is: _____________

Solution

Let the quadratic function be f(x) = ax²+bx+c where a,b, c are arbitrary real numbers. Then df/dx = 2ax+b and d²f/dx² = 2a. Since the maximum value of f(x) is 15 at x = -14, 15 = a(-14)² -14b +c or, 196 a -14b +c = 15…(1) Also, df/dx = 0 when x = -14 so that 2a(-14)+b = 0 or, b = 28a…(2). Then the 1^st equation changes to 196a -14 (28a) +c = 15 or, 196a-392a +c = 15 or, c = 15+ 196a…(3) and f(x) = ax²+28ax+15+ 196a = a(x² + 28x+ 196) +15 = a(x+14)² +15. This is the equation of a parabola with vertex at (-14,15). Further, d²f/dx² = 2a has to be negative so that a is negative. Thus, the parabola opens downwards.

When f(x) = 0, we have a(x+14)² +15 = 0 or, (x+14)² = -15/a or, x+14 = ±(-15/a) or, x = -14±(-15/a).

Thus,the domain of f is [-14-(-15/a), -14+(-15/a)] where a is an arbitrary finite real number.

The range of f is ( -, 15]