Vertex Form of a Quadratic Function
SAT Math — Algebra & Advanced Problem-Solving
Key Equation: $y = a(x - h)^2 + k$ where the vertex is $(h, k)$
1. Introduction — What Is Vertex Form?
1.1 Definition
Vertex form is a specific way to write a quadratic function that reveals the vertex (the turning point) directly from the equation:
$$y = a(x - h)^2 + k$$
- $a$ — determines the direction (up/down) and width of the parabola
- $h$ — the x-coordinate of the vertex (horizontal shift)
- $k$ — the y-coordinate of the vertex (vertical shift)
1.2 Why Vertex Form Matters on the SAT
✅ Shows vertex immediately
No calculation needed — just read $(h, k)$.
✅ Easy to graph
Plot the vertex, then use $a$ for shape.
✅ Shows transformations
Shifts, stretches, and reflections from $y = x^2$.
1.3 Three Forms of a Quadratic — Comparison
| Form | Equation | Reveals Directly |
|---|---|---|
| Standard Form | $y = ax^2 + bx + c$ | $y$-intercept $(0, c)$ |
| Vertex Form | $y = a(x - h)^2 + k$ | Vertex $(h, k)$ |
| Factored Form | $y = a(x - r_1)(x - r_2)$ | $x$-intercepts $r_1, r_2$ |
Each form has its own advantages. The SAT frequently asks you to convert between forms or read information from a given form.
2. Structure of Vertex Form — Parameters $a$, $h$, $k$
⚠️ CRITICAL SIGN DETAIL (Most Common SAT Trap!)
The equation uses $(x \mathbf{-} h)$, so $h$ has the OPPOSITE sign of what appears inside the parentheses.
- $(x - 3)^2$ → $h = +3$
- $(x + 4)^2 = (x - (-4))^2$ → $h = -4$
Quick-Read Examples
| Equation | $a$ | $h$ | $k$ | Vertex $(h,k)$ |
|---|---|---|---|---|
| $y = (x - 3)^2 + 5$ | $1$ | $3$ | $5$ | $(3, 5)$ |
| $y = (x + 4)^2 - 2$ | $1$ | $-4$ | $-2$ | $(-4, -2)$ |
| $y = 2(x - 1)^2 + 7$ | $2$ | $1$ | $7$ | $(1, 7)$ |
| $y = -(x + 5)^2 - 3$ | $-1$ | $-5$ | $-3$ | $(-5, -3)$ |
| $y = 3(x - 5)^2$ | $3$ | $5$ | $0$ | $(5, 0)$ |
| $y = x^2 + 3$ | $1$ | $0$ | $3$ | $(0, 3)$ |
| $y = -2(x + 1)^2 + 6$ | $-2$ | $-1$ | $6$ | $(-1, 6)$ |
When $h = 0$: $y = ax^2 + k$ — vertex on the $y$-axis at $(0, k)$.
When $k = 0$: $y = a(x-h)^2$ — vertex on the $x$-axis at $(h, 0)$.
Both $0$: $y = ax^2$ — vertex at the origin $(0, 0)$.