# 5 Ways to Find the Vertex

There are several math functions that use vertices. Polyhedra have them, systems of inequalities can have one or more vertices, and parables or quadratic equations can also have them. Finding the vertex varies by situation, but here are guidelines you should be aware of in each scenario.

## Steps

### Method 1 of 5: Finding the Number of Vertices in a Polygon

#### Step 1. Learn Euler's formula

Euler's formula, as used in reference to geometry and graphics, states that for any non-intersecting polyhedron, the number of faces plus the number of vertices minus the number of edges will always equal 2.

• Written as an equation, the formula can be defined as: F + V - E = 2

• F refers to the number of faces.
• V refers to the number of vertices, or corners.
• And it refers to the number of edges.

#### Step 2. Rearrange the formula to find the number of vertices

If you know how many faces and edges a polyhedron has, you can quickly count the number of vertices using Euler's formula. Subtract F from both sides of the equation and add E to both, isolating V from the other.

• V = 2 - F + E

#### Step 3. Enter the numbers and solve the equation

All you have to do at this point is put the sides and edge numbers into the equation before adding or subtracting. The answer you get will tell you the number of vertices and complete the problem.

• Example: A polyhedron has 6 faces and 12 edges.

• V = 2 - F + E
• V = 2 - 6 + 12
• V = -4 + 12
• V = 8

### Method 2 of 5: Discovering Vertices in Linear Inequality Systems

#### Step 1. Graph the solutions of the linear inequalities system

In some cases, graphing the solutions of all inequalities can visually show you where some, if not all, of the vertices will be. However, when it doesn't, you will need to find it algebraically.

### If you're using a graphing calculator, it's usually possible to scroll to the vertices and find the coordinates that way

#### Step 2. Transform inequalities into equations

To solve the system of inequalities, you will need to temporarily transform the inequalities into equations, allowing you the ability to find the values ​​of x and y.

• Example: In the following inequality system:

• y < x
• y > -x + 4
• Transform inequalities into:

• y = x
• y = -x + 4

#### Step 3. Replace one variable with another

Although there are a few different ways you can resolve to x and y, replacement is often the easiest to use. Enter the value of y from one equation to the other, effectively "replacing" y on the other with the values x additional.

• Example: If:

• y = x
• y = -x + 4
• Then, y = -x + 4 can be written as:

### x = -x + 4

#### Step 4. Solve for the first variable

Now that you only have one variable in the equation, you can easily solve for that variable, x, as you would any other: adding, subtracting, dividing and multiplying.

• Example: x = -x + 4

• x + x = -x + x + 4
• 2x = 4
• 2x / 2 = 4 / 2
• x = 2

#### Step 5. Solve for the remaining variable

Enter the new value for x in one of the original equations to find the value of y.

• Example: y = x

### y = 2

#### Step 6. Determine the vertex

The vertex is simply the coordinate consisting of your new values. x and y.

### Method 3 of 5: Finding the Vertex of a Parabola with Symmetry Axes

#### Step 1. Factor the equation

Rewrite the quadratic equation in its factored form. There are several ways to factor a quadratic equation, but when you do, you'll be left with two sets in parentheses that, multiplied, equal the original equation.

• Example (through decomposition):

• 3x2 - 6x - 45
• Find the common factor: 3 (x2 - 2x - 15)
• Multiply the terms a and c: 1 × -15 = -15
• Find two numbers with a product equal to -15 and a sum equal to the value b, -2: 3 × -5 = -15; 3 - 5 = -2
• Substitute the two values ​​into the equation: ax2 + kx + hx + c: 3 (x2 + 3x - 5x - 15)
• Factor the polynomial by grouping: f(x) = 3 × (x + 3) × (x - 5)

#### Step 2. Find the point at which the equation crosses the x-axis

Whenever the function of x, or f(x), is equal to 0, the parabola will cross the x-axis. This will happen when any of the sets of factors equals 0.

• Example: x + 3; -3 + 3 = 0

• x - 5; 5 - 5 = 0
• Therefore, the roots are: (-3, 0) and (5, 0)

#### Step 3. Calculate the midpoint

The axis of symmetry of the equation will be directly between the two roots of the equation. You will need to find the axis of symmetry since the vertex is on top of it.

### Example: x = 1; this value is directly between -3 and 5

#### Step 4. Put the value of x into the original equation

Put the value of x for the axis of symmetry into any of the equations for the parabola. The y value will be the y value for the vertex.

• Example: y = 3x2 - 6x - 45 = 3(1)2 - 6(1) - 45 = -48

#### Step 5. Write the vertex point

At this point, the last values ​​for x and y should give you the vertex coordinates.

### Method 4 of 5: Finding the Vertex of a Parabola Completing the Square

#### Step 1. Rewrite the original equation in its vertex form

The "vertex" shape of an equation is written as y = a(x - h)2 + k, and the vertex will be (h, k). Your current quadratic equation will need to be rewritten in this form, and to do this you must complete the square.

• Example: y = -x2 - 8x - 15

#### Step 2. Isolate the a value

Factor the coefficient of the first term, a, from the first two terms of the equation. Leave the final term, c, for now.

• Example: -1 (x2 + 8x) - 15

#### Step 3. Find a third term for the parentheses

The third term must complete the set in parentheses so that the values ​​between them form a perfect square. This new term will be the squared value of half the coefficient of the central term.

• Example: 8 / 2 = 4; 4 × 4 = 16; soon,

• -1 (x2 + 8x + 16)
• Also, remember that what you do internally must be done externally:

• y = -1 (x2 + 8x + 16) - 15 + 16

#### Step 4. Simplify the equation

Since the parentheses now form a perfect square, you can simplify the parenthetical portion to the factored form. Simultaneously, it is possible to perform necessary additions or subtractions to values ​​outside the parentheses.

• Example: y = -1 (x + 4)2 + 1

#### Step 5. Find out which coordinates are based on the vertex equation

Remember that the vertex shape of an equation is given by y = a(x - h)2 + k, with (h, k) representing the coordinates of the vertex. You now have enough information to enter the values ​​in the h and k spaces and complete the problem.

• k = 1
• h = -4
• Therefore, the vertex of this equation can be found in: (-4, 1)

### Method 5 of 5: Finding the Vertex of a Parabola with a Simple Formula

#### Step 1. Find the x coordinate of the vertex directly

If the equation of your parable can be written as y = ax2 + bx + c, the x of the vertex can be discovered through the formula x = -b / 2a. Simply input a and b values ​​from the equation to find x.

• Example: y = -x2 - 8x - 15
• x = -b / 2a = -(-8) / 2 × (-1) = 8 / (-2) = -4
• x = -4

#### Step 2. Enter this value into the original equation

By entering the value of x into the equation, you can solve for y. This y value will be the y coordinate of your vertex.

• Example: y = -x2 - 8x - 15 = -(-4)2 - 8(-4) - 15 = -(16) - (-32) - 15 = -16 + 32 - 15 = 1

### y = 1

#### Step 3. Write the coordinates of the vertex

The x and y values ​​obtained will be the coordinates of its vertex point.