Writing a quadratic as $y = a(x - h)^2 + k$ shows its vertex at a glance. This is the transformation idea applied to $y = x^2$: shift it right $h$, stretch by $a$, shift it up $k$. It makes the features of a parabola easy to read.
Why it works
A square is never negative, so $(x - h)^2 \ge 0$, and it equals $0$ only when $x = h$. If $a > 0$, then $a(x - h)^2 \ge 0$ too, so $y \ge k$, with $y = k$ exactly at $x = h$: the lowest point is $(h, k)$. If $a < 0$ the inequality flips and $(h, k)$ is the highest point. Either way the vertex is $(h, k)$ and the axis of symmetry is $x = h$.
Key idea
In $y = a(x - h)^2 + k$ the vertex is $(h, k)$: watch the sign, since $x + 3 = x - (-3)$ gives $h = -3$. The axis is $x = h$, and $k$ is the minimum ($a > 0$) or maximum ($a < 0$) value.
Worked examples
Worked example: Read the vertex
Find the vertex and the maximum or minimum of $y = -2(x + 3)^2 + 5$.
- $x + 3 = x - (-3)$, so $h = -3$; $k = 5$. Vertex $(-3, 5)$.
- $a = -2 < 0$: opens down, so $5$ is the maximum value.
Answer: Vertex $(-3, 5)$; maximum $5$.
Worked example: Write from a vertex and a point
A parabola has vertex $(1, -4)$ and passes through $(3, 4)$. Write its equation.
- $y = a(x - 1)^2 - 4$.
- Use the point: $4 = a(3 - 1)^2 - 4 = 4a - 4$, so $4a = 8$ and $a = 2$.
Answer: $y = 2(x - 1)^2 - 4$
Common mistake
Reading the vertex of $y = (x - 2)^2 + 3$ as $(-2, 3)$. The form subtracts $h$, so $x - 2$ means $h = +2$. Check: at $x = 2$ the square is $0$ and $y = 3$, the smallest possible value, so $(2, 3)$ is the vertex.