Vertex form of a parabola

You know the parabola as a graceful curve, but from an equation like $y = x^2 - 6x + 11$ its highest or lowest point is hard to spot. There is a second way of writing it that reveals exactly this point at once: the vertex form. Once you have it, there is nothing left to compute, you simply read the vertex off.

What the vertex form looks like

In vertex form the parabola reads $y = a(x - d)^2 + e$. You already know the number $a$ from $y = a x^2 + c$: it stretches or squeezes the curve and decides whether it opens upwards or downwards. What is new are $d$ and $e$: together they form the vertex $(d, e)$. You need no more than this to know where the parabola turns. Our example from the start reads $y = (x - 3)^2 + 2$ in vertex form, the very same parabola as $y = x^2 - 6x + 11$, just written differently. So its vertex sits at $(3, 2)$.

Why a minus in front of d?

Watch the sign: in $y = (x - 3)^2$ there is a minus, and the vertex is at $x = 3$, so to the right. In $y = (x + 4)^2$ there is a plus, which you may read as $x - (-4)$, and the vertex is at $x = -4$, so to the left. The term becomes zero when the bracket becomes zero, and that is exactly where the vertex sits. The value $e$ behind it lifts the parabola up by $e$ or lowers it.

Shifting instead of redrawing

The vertex form tells a story in two steps: start with the basic parabola $y = a x^2$, whose vertex sits at the origin. Then you shift it $d$ to the right and $e$ upwards. For $y = 2(x - 1)^2 + 3$ you take the narrow parabola $y = 2x^2$ and carry its vertex to $(1, 3)$. If $d$ or $e$ is negative, the shift goes the other way, to the left or downwards. That turns a formula into a movement you can picture.