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Vertex Form: Equation & Functions

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  • 0:00 Vertex Form
  • 0:33 Vertex of a Quadratic Function
  • 1:49 Converting to and from…
  • 3:38 From Vertex to Graph
  • 5:28 Lesson Summary
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Lesson Transcript
Instructor
Glenda Boozer
Expert Contributor
Simona Hodis

Simona received her PhD in Applied Mathematics in 2010 and is a college professor teaching undergraduate mathematics courses.

Just as y = mx + b is a useful format for graphing linear functions, y = a(x - h)^2 + k is a useful format for graphing quadratic functions. We will explore its uses and learn how to convert any quadratic equation into vertex form.

Vertex Form

A quadratic equation can be written as ax2 + bx + c = 0, as long as a does not equal 0. We usually see this form, known as standard form, because it is used for more than one method of solving quadratic equations. In this lesson, however, we will look at vertex form, which is written as a(x - h)2 + k. We can make good use of this format for graphing quadratic functions.

Vertex of a Quadratic Function

Let's graph the function associated with a quadratic equation: f'(x) = ax2 + bx + c or y = ax2 + bx + c.

Specifically, we can use y = 3x2 + 6x + 1 as an example.

y = 3x^2 + 6x + 1

Notice that the point at (-1,-2) is the lowest point on the graph. This is what we call the vertex. We can also find the coordinates of the vertex from the equation itself, using this formula: x = -b/2a.

In this case, x = -6/2(3) = -6/6 = -1. Hey, it works! Now since we know that x is -1, we can determine that y = 3(-1)2 + 6(-1) + 1 = 3(1) + (-6) + 1 = 3 - 6 + 1 = -3 + 1 = -2. So, there is our (-1,-2). This means that we do not have to graph the equation to find the vertex.

Converting to and from Vertex Form

Let's take the equation in standard form that we have already been working with:

y = 3x2 + 6x + 1

We already found the vertex to be (-1,-2).

The vertex form will look like:

a(x - h)2 + k

The a will be the same as in standard form, so we have:

y = 3(x - h)2 + k

Well, it turns out that the x-coordinate (our -1) is our h, and the y-coordinate (our -2) is the k. That gives us:

y = 3(x - (-1))2 + (-2), which is y = 3(x + 1)2 - 2

Thus, we use this method to convert from standard form to vertex form:

  • Find the x-coordinate using the formula x = -b/2a
  • Find the y-coordinate by evaluating f(x) = ax2 + bx + c with our value for x
  • Use the a from the standard form, the x-coordinate for h and the y-coordinate for k in y = a(x - h)2 + k

If we want to go the other way, we can simply multiply (x - h) by itself, multiply that result by a, and add k.

For example:

3(x + 1)2 - 2 = 3(x2 + 2x + 1) - 2 = 3x2 + 6x + 3 - 2 = 3x2 + 6x + 1

That's what we started with, so we know how to convert standard form to vertex form or vertex form back to standard form.

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Additional Activities

Vertex Form - Optimization Problems


Review Topics

The vertex form of a quadratic function is an expression that easily provides the coordinates of the vertex point on the parabola.

The vertex form is

The vertex coordinates are

The vertex point is the extreme point on a parabola. If the quadratic term is positive, the parabola opens up, therefore the vertex is a minimum point.

And when the quadratic term is negative, the parabola opens down, and the vertex is a maximum point.

Optimization Problems

Quadratic functions often appear in optimization problems where we need to obtain the extreme value of the function or the coordinates of the vertex.

For example, to find the dimensions of a rectangle when a given perimeter and largest area are obtained by finding the x-coordinate of the vertex of a quadratic equation.

In general, the optimization problems simplify to finding a function whose extreme values are to be determined by imposing a given constraint.

To practice this technique, let's do the following problem.

Given a rectangular field that we want to fence with 200 yards, we will find the dimensions of the rectangular filed of the largest area if only three sides need to be fenced (the fourth side is along an existing wall).

To find the dimensions, perform the following steps.

1. Denoting the side of the field parallel to the wall with x and the other side with y, obtain a function of the area of the field in terms of x and y.

2. Provide an expression for the constraint.

3. Write the function obtained in 1. in terms of a single variable, using the constraint in 2.

4. Give the x- coordinate of the vertex corresponding to the quadratic function from 3.

Solutions

1. A = xy

2. 2x + y = 200

3. A(x) = x(200 - 2x).

4. x = 50 yards, using a = -2, b = 200 to obtain the x-coordinate of the vertex.

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