How to Write Quadratic Functions Video

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  • 0:05 A Quadratic Function
  • 0:41 Given a Graph
  • 2:52 Given Three Points
  • 7:14 Example
  • 8:39 Lesson Summary
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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.

A Quadratic Function

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.

Given a Graph

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.

Given Three Points

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:

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