Show that for a quadratic function y = ax2 + bx + c with a < 0, the function has a maximum value at x = -b/(2a) using the method of completing the square.

To do this, add and subtract $\left(\frac{b}{2a}\right)^2$.

$$y = a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c$$ $$y = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c$$

$$y = a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b^2}{4a^2}\right) + c$$ $$y = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$

$$y = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}$$

This is the vertex form $y = a(x-h)^2 + k$, where the vertex is at $(h, k)$. In this case, $h = -\frac{b}{2a}$ and $k = \frac{4ac - b^2}{4a}$.

The term $\left(x + \frac{b}{2a}\right)^2$ is always greater than or equal to zero, i.e., $\left(x + \frac{b}{2a}\right)^2 \ge 0$.
The minimum value of this term is 0, which occurs when $x + \frac{b}{2a} = 0$, or $x = -\frac{b}{2a}$.

We are given that $a < 0$.
Since $a$ is negative and $\left(x + \frac{b}{2a}\right)^2$ is non-negative, the product $a\left(x + \frac{b}{2a}\right)^2$ will always be less than or equal to zero:

$$a\left(x + \frac{b}{2a}\right)^2 \le 0$$

To maximize $y$, we need to make the term $a\left(x + \frac{b}{2a}\right)^2$ as large as possible. Since it is always non-positive, its maximum value is 0.
This occurs when $\left(x + \frac{b}{2a}\right)^2 = 0$, which means $x = -\frac{b}{2a}$.

At $x = -\frac{b}{2a}$, the function $y$ reaches its maximum value:

$$y_{max} = a(0) + \frac{4ac - b^2}{4a} = \frac{4ac - b^2}{4a}$$

Therefore, if $a < 0$, the quadratic function $y = ax^2 + bx + c$ has a maximum value when $x = \boxed{-\frac{b}{2a}}$.