Understanding the Vertex of a Parabola: A Comprehensive Guide

When examining the graph of a parabola defined by a quadratic equation, identifying the vertex is a fundamental step. In this guide, we explore how to find the vertex for the given equation y 6x^2 - 5, revealing both the process and application.

Step-by-Step Guide on Finding the Vertex

The vertex form of a quadratic equation is given by:

y a(x - h)^2 k

In this form, the vertex of the parabola is at the point (h, k).

Application to the Given Equation

Given Equation

The given equation is y 6x^2 - 5

Step 1: Rewrite the Equation

The term x^2 can be rewritten as (x - 0)^2, indicating that h 0.

Step 2: Identify the Constant Term

The constant term k -5.

Thus, the vertex of the parabola is (h, k) (0, -5).

Alternative Approach

To further verify the vertex, let's use the modified equation provided:

Modified Equation

frac{1}{6}y - 5 x^2

Step 1: Isolate the Parabola Equation

First, multiply both sides by 6 to clear the fraction:

y - 30 6x^2

Step 2: Rearrange to Vertex Form

Rearrange this to the standard form of a parabola:

y - 5 6(x - 0)^2

Step 3: Identify Vertex Coordinates

Now, the vertex is clearly at (0, 5). Since we need the form y 6(x - h)^2 k, we can write:

y 6(x - (-2))^2 5

This simplifies to:

y 6(x 2)^2 5

Thus, the vertex, using this form, is (-2, 5).

Additional Insights

Focal Properties

The vertex form can also be used to determine other key properties of the parabola, such as the focus, directrix, and major axis.

Focus and Directrix Calculations

For the parabola given in vertex form y a(x - h)^2 k, the focus and directrix can be determined by:

Focus (h, k frac{1}{4a})

Directrix y k - frac{1}{4a}

For our specific parabola, y 6(x 2)^2 5 where a 6, the calculations are:

frac{1}{4a} frac{1}{24}

Focus: (-2, 5 frac{1}{24}) (-2, 5 frac{1}{24})

Directrix: y 5 - frac{1}{24} frac{119}{24}

Verification of the Vertex

A point on the parabola can be used to verify the vertex. Checking the point P(-1, 11):

Substitute into the equation:

11 6(-1 2)^2 5

This simplifies to:

11 6(1) 5 6 5 11

This confirms that P(-1, 11) is indeed on the parabola.

The distance from the directrix to the focus can be calculated to be equal:

PF sqrt{(11 - frac{119}{24})^2 (-1 2)^2} frac{145}{24}

This verifies the consistency of the parabolic properties.