Maximum and minimum of a quadratic function

The maximum or minimum of a quadratic function is found at the vertex of its graph. Let us find the maximum of $y = -x^2 + 4x - 1$.

Since the coefficient of $x^2$ is negative, this graph is a downward-opening parabola. When it opens downward, the vertex is the highest point of the graph, where the function attains its maximum (there is no lower bound, so no minimum).

We find the vertex by completing the square: $-x^2 + 4x - 1 = -(x^2 - 4x) - 1 = -\left((x - 2)^2 - 4\right) - 1 = -(x - 2)^2 + 3$, so the vertex is $(2, 3)$. Since $-(x - 2)^2$ is $-1$ times a square, it is always at most $0$, and it is largest when $x = 2$. So the maximum value $3$ is taken at $x = 2$.

In general $y = a(x - p)^2 + q$ has, when $a < 0$ (opening downward), a maximum $q$ at $x = p$, and when $a > 0$ (opening upward), a minimum $q$ at $x = p$. Whether it is a maximum or a minimum depends only on the sign of the coefficient of $x^2$. The large dot on the graph is the vertex $(2, 3)$, where the maximum is attained.