In Algebra 1 you found the vertex by completing the square. That works every time, but it is slow. Doing it once with letters instead of numbers gives a formula you can use directly.
Key idea
The vertex of $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Substitute that $x$ back into the function to get the $y$-coordinate. The axis of symmetry is the line $x = -\frac{b}{2a}$, and the vertex height is the minimum value when $a > 0$ and the maximum when $a < 0$.
Why it works
Complete the square on $ax^2 + bx + c$ with letters. Factor $a$ out of the first two terms: $a\left(x^2 + \frac{b}{a}x\right) + c$. Half of $\frac{b}{a}$ is $\frac{b}{2a}$, so $$x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}.$$ The squared part is smallest (zero) when $x + \frac{b}{2a} = 0$, that is, at $x = -\frac{b}{2a}$. There is another way to see it: the zeros of $ax^2 + bx$ are $0$ and $-\frac{b}{a}$, and the vertex sits halfway between them. Adding $c$ moves the parabola up or down but not sideways.
Worked examples
Worked example: A vertex with whole numbers
Find the vertex of $y = 2x^2 - 12x + 7$.
- $a = 2$ and $b = -12$, so $x = -\frac{-12}{2 \cdot 2} = \frac{12}{4} = 3$.
- $y = 2(3)^2 - 12(3) + 7 = 18 - 36 + 7$ $= -11$.
Answer: $(3, -11)$
Worked example: A maximum value
A ball's height is $h(t) = -16t^2 + 48t + 5$ feet after $t$ seconds. What is its greatest height?
- $a = -16 < 0$, so the parabola opens down and the vertex is the highest point.
- $t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5$ seconds.
- $h(1.5) = -16(2.25) + 72 + 5$ $= -36 + 77 = 41$.
Answer: 41 feet, at $t = 1.5$ seconds
Worked example: A fraction for the vertex
Find the vertex of $f(x) = x^2 + 3x + 1$.
- $x = -\frac{3}{2 \cdot 1} = -\frac{3}{2}$.
- Substitute: $$f\left(-\tfrac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} + 1 = \frac{9}{4} - \frac{18}{4} + \frac{4}{4} = -\frac{5}{4}.$$
Answer: $\left(-\frac{3}{2}, -\frac{5}{4}\right)$
Common mistake
Dropping the negative sign and using $x = \frac{b}{2a}$. For $y = x^2 - 6x$ that puts the vertex at $x = -3$ instead of $x = 3$. Check: the zeros of $x^2 - 6x$ are 0 and 6, and the vertex must be halfway between them, at $x = 3$.