Parabolas and Vertex Form — Mathematics Knowledge System

Overview

A parabola is the graph of a quadratic function f(x) = ax^2 + bx + c. The vertex form f(x) = a(x - h)^2 + k makes the vertex (h, k) explicit and the axis of symmetry x = h immediate. The parameter a controls opening direction (a > 0: opens up, a < 0: opens down) and width (|a| larger means narrower).

Intuition

The vertex form f(x) = a(x-h)^2 + k is a transformation of f(x) = ax^2. The parabola is shifted right by h, up by k. The vertex (h,k) is the minimum (a>0) or maximum (a<0). Converting: complete the square on ax^2 + bx + c to get a(x - (-b/2a))^2 + (c - b^2/4a). So vertex is at x = -b/(2a).

Formal Definition

Definition

Vertex form: f(x) = a(x - h)^2 + k, vertex (h, k), axis of symmetry x = h. Standard form: f(x) = ax^2 + bx + c. Vertex from standard: h = -b/(2a), k = f(h) = c - b^2/(4a). The parabola opens up if a > 0, down if a < 0. Width: larger |a| gives narrower parabola (steeper sides).

f(x) = a(x - h)^2 + k

Vertex form

h = -\frac{b}{2a},\quad k = c - \frac{b^2}{4a}

Vertex coordinates

x = h \text{ (axis of symmetry)}

Axis of symmetry

Notation

Notation	Meaning
$(h, k)$	Vertex of the parabola
$a$	Leading coefficient; controls opening and width

Theorems

Theorem 1: Vertex is Extremum

For f(x) = a(x-h)^2 + k: if a > 0, the vertex (h,k) is the global minimum of f. If a < 0, the vertex is the global maximum. There are no other local extrema.

Worked Examples

1
Complete the square: x^2 - 6x = (x-3)^2 - 9.
$f(x) = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$
2
Vertex is (h, k) = (3, -4).
$\text{Vertex: } (3, -4)$

✓ Answer

f(x) = (x-3)^2 - 4, vertex at (3, -4).

Practice Problems

Easyfill in blank

Find the vertex of f(x) = 2x^2 + 8x + 3.

Common Mistakes

Common Mistake

Reading vertex as (h, k) = (2, 5) from f(x) = (x-2)^2 + 5, forgetting the minus sign.

Vertex is at x = h, where (x - h) = 0, so h is the number subtracted. f(x) = (x-2)^2 + 5 has vertex (2, 5). f(x) = (x+2)^2 + 5 has vertex (-2, 5).

Quiz

The parabola f(x) = -3(x+2)^2 + 5 opens:

UpwardDownwardLeftRight

Historical Background

Parabolas were studied by Greek mathematicians as conic sections -- curves formed by slicing a cone. Apollonius of Perga (c. 200 BCE) named the parabola and studied its properties. Galileo discovered (c. 1610) that projectile trajectories are parabolic. The connection between geometric parabolas and quadratic equations was made explicit through Descartes' coordinate geometry.

200 BCE
Apollonius names and classifies conic sections including parabolas
Apollonius of Perga
1610
Galileo discovers projectile motion follows a parabolic path
Galileo Galilei

Summary

Vertex form: f(x) = a(x-h)^2 + k, vertex (h,k), axis x = h.
a > 0: opens up (minimum at vertex). a < 0: opens down (maximum at vertex).
Convert standard to vertex by completing the square: h = -b/(2a), k = f(h).

References

BookHall, B. and Fabricant, M. Algebra 1. Prentice Hall, 2001.