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Holt McDougal Algebra 2: Online Textbook Help14 chapters | 233 lessons

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

In the real world, you won't always be given equations as you are in textbooks. Often, you'll need to find equations using a few points or other pieces of data. Learn how to find and write quadratic equations.

This lesson is about writing quadratic functions. A **quadratic function** is a polynomial function of degree 2. So, y = x^2 is a quadratic equation, as is y = 3x^2 + x + 1. All of these are polynomial functions of degree 2. Often you will be given the quadratic function for a particular problem. You'll be given several data points, or a graph approximating all the data points combined. It's then your job to come up with a quadratic function that fits the provided information. Let's go ahead and see how we can do this.

Let's begin with writing the quadratic function given a graph. Let's say we are given this graph to work with. Before we can begin to write our quadratic function, we need to figure out the location of the vertex, or tip of the quadratic, along with one other point. Looking at our graph, we see that our vertex, our tip, is at (0, -1). The other point that we can easily see is (0.5, 0). Having found these two points, we can now proceed. We will use the vertex form of the quadratic function, which is:

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

Where a is a constant and h and k are the x and y values of the vertex (h,k). Remember, our 'f(x)' is the same as 'y'. The steps are as follows:

1. Plug the vertex values of h and k into the vertex form of the quadratic function and simplify

2. Now, plug in the x and y values of the other point into the simplified quadratic function from step 1

3. Solve for a

4. Now, plug in the a value into the simplified quadratic function from step 1. This is your quadratic function

Let's try it now: our vertex is (0, -1) and our other point is (0.5, 0). This means that our h = 0, k= -1, x = 0.5, y = 0. So we begin by plugging in our h and k values. We get this:

Y = a(x-0)^2 - 1

Simplifying this we get:

Y = ax^2 - 1

Now we plug in our x and y; remember that our f(x) is the same as y. So we get:

0 = a(0.5)^2 - 1

Solving for a, we get this:

0 = a(0.25) - 1

1 = 0.25a

4 = a

Plugging this a value into our function, y = ax^2 - 1, we get:

Y = 4x^2 - 1

This function then is our answer.

Now let's see what the steps are when we are given three points instead of the graph. Say our problem is like this: write the quadratic function for the graph that passes through the points (-2,1), (-4,4), and (2,1). To write the quadratic function for this problem, we need to use the general form of the quadratic function, which is:

F(x) = ax^2 + bx + c

Where a, b, and c are constants. Again, remember that our f(x) is the same as y. What we are going to do, is to create a system of three equations that we will use to solve for our three unknown constants, a, b, and c. The steps are as follows:

1. Plug in the x and y values into the general form of the quadratic function and simplify. So this for all three points so you get one equation for each point. You should have three equations.

2. Solve this system of three equations with either substitution or elimination.

3. Plug your found values for a,b, and c into the general form of the quadratic function. Simplify to find your quadratic function that passes through all three of the given points.

Let's try it out! Our three points are (-2,1), (-4,4) and (2,1). Plugging these points into the general form and simplifying, we get these three equations:

For the point (-2,1) we have 1 = a(-2)^2 + b(-2) + c, which simplifies to:

1= 4a - 2b +c

For the point (-4,4), we have 4 = a(-4)^2 + b(-4) + c, which simplifies to:

4 = 16a - 4b +c

For the point (2,1), 1 = a(2)^2 + b(2) + c, which simplifies to:

1 = 4a + 2b + c

Now, taking these three equations and solving the system by elimination, we get this:

We can subtract 1 = 4a - 2b + c from 4 = 16a - 4b + c to eliminate the c variable. We get:

4 = 16a - 4b + c - 1 = 4a - 2b + c =

3= 12a - 2b

Using a different pair of equations, we will again eliminate the c variable. This time we will subtract:

1 = 4a + 2b + c from 1 = 4a - 2b + c. We get:

1 = 4a - 2b + c - 1 = 4a + 2b + c =

(0 = 0a - 4b) This simplifies to:

0 = -4b

Since the last equation that we got 0 = -4b only has one variable, we can go ahead and solve this equation for that variable. If this equation had two variables, we would then use the other equation, 3 = 12a - 2b in combination with this equation to help us solve for a and b.

Since this isn't the case, we can go ahead and solve 0 = -4b for the b variable. We get:

b = 0

We now know what b equals. We can now plug this value into the 3 = 12a - 2b equation and solve for a. We get:

3 = 12a - 2(0)

3 = 12a

a = 3/12

a = ¼

We now have a = ¼ and b = 0. We can now use these two values and plug them into any one of the three equations we first wrote. Let's plug these a and b values into the equation:

1 = 4a + 2b + c

We get:

1 = 4 (1/4) + 2(0) + c

1 = 1 + c

c = 0

We have now solved for all of our variables. Our a = ¼, our b = 0, and our c = 0. Plugging these back into our general form and simplifying, we get our quadratic function:

Y = ax^2 + bx +c

Y = 1/4x^2 + 0(x) +0

Y= 1/4x^2

Let's look at one more example. Write the quadratic function for the graph that passes through the points (-1,0), (0,-1), and (1,0), where (0,-1) is the vertex. Reading through this problem, we see that we are given three points and it also tells us that -1 is the vertex. Hmmmâ€¦we can actually use either method. Let's do the method that uses the vertex form of the quadratic function. It has less steps in it and will be shorter and quicker to solve.

Since our vertex is (0,-1), our h = 0, and our k = -1, we can use any other point for our second point. Let's choose (1,0). So our x = 1 and our y = 0. So we plug in our h and k first into the vertex form and then simplify. We get:

Y = a(x-0)^2 - 1. Simplifying this we get:

Y = a(x)^2 - 1

Now plugging in 1 for x and 0 for y, we can solve for a.

0 = a(1)^2 - 1

0 = a - 1

a = 1

Now that we have a, we can plug this a value into y = a(x)^2 - 1 to find our quadratic function, we get this:

Y = (1)x^2 - 1

Y = x^2 - 1

And, we are done!

Let's review what we've learned! A **quadratic function** is a polynomial function of degree 2. The vertex form of the quadratic function is:

F(x) = a(x - h)^2 + k , where a is a constant and h and k are the x and y values of the vertex (h,k).

To write the quadratic function when you know the vertex and one other point, follow these steps:

1. Plug the vertex values of h and k into the vertex form of the quadratic function and simplify

2. Now plug in the x and y values of the other point into the simplified quadratic function from step 1

3. Solve for a

4. Now, plug in the a value into the simplified quadratic function from step 1. This is your quadratic function.

The general form of the quadratic function is:

F(x) = ax^2 + bx + c, where a, b, and c are constants.

To write the quadratic function when you are given three points, follow these steps:

1. Plug in the x and y values into the general form of the quadratic function and simplify. Do this for all three points so you get one equation for each point. You should have three equations.

2. Solve this system of three equations with either substitution or elimination.

3. Plug in your found values for a, b, and c into the general form of the quadratic function. Simplify to find our quadratic function that passes through all three of the given points.

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15 in chapter 5 of the course:

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Holt McDougal Algebra 2: Online Textbook Help14 chapters | 233 lessons

- What is a Parabola? 4:36
- How to Reflect Quadratic Equations 3:48
- Parabolas in Standard, Intercept, and Vertex Form 6:15
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