Graphing Quadratics — Vertex Form
Vertex Form of a Parabola
Every quadratic function y = ax² + bx + c graphs as a parabola — a smooth U-shaped curve that opens upward when a is positive and downward when a is negative. The vertex of the parabola is its highest or lowest point, making it essential for optimization problems.
The vertex form y = a(x − h)² + k makes the vertex immediately visible: it sits at the point (h, k). Converting between standard and vertex form, often by completing the square, is a key skill for both graphing and problem solving.
This lesson covers graphing parabolas, finding the vertex, identifying the axis of symmetry, and determining maximum or minimum values.
$$f(x) = a(x - h)^2 + k$$
Vertex: $(h, k)$. Axis of symmetry: $x = h$.
- $a > 0$: opens up (vertex is minimum)
- $a < 0$: opens down (vertex is maximum)
- $|a| > 1$: narrower than $y = x^2$
- $|a| < 1$: wider than $y = x^2$
Given $f(x) = ax^2 + bx + c$:
$$h = -\frac{b}{2a}, \quad k = f(h)$$
Find the vertex of $f(x) = 2x^2 - 12x + 22$.
$h = -\dfrac{-12}{2(2)} = 3$. $k = 2(9) - 36 + 22 = 4$.
Vertex: $(3, 4)$. Vertex form: $f(x) = 2(x - 3)^2 + 4$.
Identify vertex, direction, axis, and intercepts of $f(x) = -(x + 1)^2 + 9$.
Vertex: $(-1, 9)$. Opens down ($a = -1$). Axis: $x = -1$.
$y$-int: $f(0) = -1 + 9 = 8$.
$x$-int: $0 = -(x+1)^2 + 9 \Rightarrow (x+1)^2 = 9 \Rightarrow x = 2$ or $x = -4$.
$f(x) = x^2 + 4x + 7$. Find vertex, direction, and range.
$h = -2$. $k = 4 - 8 + 7 = 3$. Vertex: $(-2, 3)$. Opens up.
Range: $[3, \infty)$. No $x$-intercepts ($\Delta = 16 - 28 < 0$).
Graphing Checklist
- Find vertex $(h, k)$.
- Draw axis of symmetry $x = h$.
- Find $y$-intercept: $f(0)$.
- Find $x$-intercepts (if real).
- Plot symmetric points and sketch.
Practice Problems
Show Answer Key
1. $(3, -4)$
2. $(-3, 7)$
3. Down ($a = -1 < 0$)
4. $f(0) = 3(1) - 4 = -1$
5. $x = 5$ and $x = -1$
6. $x = -2$
7. $(x - 5)^2 - 4$
8. $(-\infty, 5]$
9. Min value at $x = 2$: $f(2) = 12 - 24 + 7 = -5$
10. $\Delta = 4 - 20 = -16 < 0$. No.