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High School Precalculus: Tutoring Solution32 chapters | 265 lessons

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Lesson Transcript

Instructor:
*Glenda Boozer*

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.

A quadratic equation can be written as *ax**2* + *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.

Let's graph the function associated with a quadratic equation: *f'(x)* = *ax**2* + *bx* + *c* or *y* = *ax**2* + *bx* + *c*.

Specifically, we can use *y* = 3*x**2* + 6*x* + 1 as an example.

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*/2*a*.

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.

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

*y* = 3*x**2* + 6*x* + 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*/2*a* - Find the y-coordinate by evaluating
*f(x)*=*ax**2*+*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(*x**2* + 2*x* + 1) - 2 = 3*x**2* + 6*x* + 3 - 2 = 3x*2* + 6*x* + 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.

Now we have a vertex form. If we want to graph it, we can set things up in a systematic way:

- Use the vertex as one of the points to plot on the graph
- Try using a few
*x*values that are less than the*x*-coordinate of the vertex and a few values that are greater than the*x*-coordinate of the vertex - Find the
*y*-coordinate to go with each value of*x* - Plot the points, or enter them into a graphing program

Let's try it with a different example, like:

*y*= 2(*x*- 3)*2* + 1

Since in this equation, 3 is *h* and 1 is *k*, the vertex is (3,1). Let's choose several values for *x*: 0, 1, 2, 3, 4, 5, 6.

If *x* is 0, then *y* = 2(0 - 3)*2* + 1 = 2(-3)*2* + 1 = 2(9) + 1 = 18 + 1 = 19

If *x* is 1, then *y* = 2(1 - 3)*2* + 1 = 2(-2)*2* + 1 = 2(4) + 1 = 8 + 1 = 9

If *'x* is 2, then *y* = 2(2 - 3)*2* + 1 = 2(-1)*2* + 1 = 2(1) + 1 = 2 + 1 = 3

If *x* is 3, we already know that *y* is 1

If *x* is 4, then *y* = 2(4 - 3)*2* + 1 = 2(1)*2* + 1 = 2 + 1 = 3

Hey, that's just like the answer for *x* = 2. One less than the *x*-value of the vertex gives the same answer as one more than the *x*-value of the vertex.

If *x* is 5, then *y* = 2(5 - 3)*2* + 1 = 2(2)*2* + 1 = 2(4) + 1 = 8 + 1 = 9

See the pattern?

If *x* is 6, then *y* = 2(6 - 3)*2* + 1 = 2(3)*2* + 1 = 2(9) + 1 = 18 + 1 = 19

Here's how it looks when we plot the points:

The **vertex form** of a quadratic equation makes graphing relatively easy. If we are working with the graph or the **vertex**, this is the format to use. We can convert between vertex form and standard form with relative ease using the formula: *x* = *-b/*2*a*.

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High School Precalculus: Tutoring Solution32 chapters | 265 lessons

- What is a Quadratic Equation? - Definition & Examples 5:13
- Solving Quadratics: Assigning the Greatest Common Factor and Multiplication Property of Zero 5:24
- How to Use the Quadratic Formula to Solve a Quadratic Equation 9:20
- How to Solve Quadratics That Are Not in Standard Form 6:14
- Solving Quadratic Inequalities Using Two Binomials 5:36
- Solving Quadratic Equations by Substitution 5:45
- Vertex Form: Equation & Functions 5:51
- Go to Introduction to Quadratics in Precalculus: Tutoring Solution

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