Modeling with Quadratic Functions

Instructor: Anna Teper-Dillard

I was born and raised in Poland. Before I ended up settling in Chicago, I traveled through Europe and Asia. I have been to China, Mongolia, Russia, Italy, Spain, Germany, Czech Republic and England. In my free time, I love reading, and being physically active. I ski, bike, swim and run long distance. I finished the Chicago Marathon, the Fox Valley Marathon, a couple of Half-Marathons and many shorter races. In the summer of 2017, I will be participating in 70.3 miles Iron Man triathlon in Michigan. Currently, I live in Woodrigde, IL with my husband and three children. I have two master degrees, one in Secondary Education and the other in Science of Mathematics. My philosophy on teaching and learning :) I am a long-life learner. Discovering new knowledge (in any subject) is fascinating and rewards us with a broader and more informative point of view. Being exposed to wider range of experiences gives us an edge in life. Although grades we earn throughout our education do not communicate the whole story about who we are, they are good indicators of many attributes and skills we have accomplished: perseverance, fortitude, hard, consistent work, collaboration, reflection on our own mistakes and yes, humbleness.

In this lesson, you will learn how to use a quadratic function in the standard, vertex, and factored forms to model a real-life scenario. You will describe and evaluate a path of a launched object using quadratic modeling.

Standard Form of Quadratic Function

Recall a quadratic function in the standard form f(x) = ax2 + bx + c, where a is the leading coefficient, b is the middle term coefficient, and c is the constant (y-intercept) of the function. Those three values can tell a lot about the behavior of the function and the real-life scenario it models. You might also remember that a graph of a quadratic function is called a parabola. Interesting fact: parabola term comes from Greek, where it literally means 'placing side by side', just like the shape of the parabola is symmetric about the line in the middle, the axis of symmetry.

Graph of a Function in Standard Form
Graph of a function in standard form

Let's start with the leading coefficient a that determines how curved the parabola is. If the a value is positive, the parabola opens up, and the function has a minimum. Similarly, if a < 0, the parabola opens down and the function will have a maximum. The minimum or the maximum value of a quadratic function is also called the vertex (h,k), where h is the x-value of the vertex and k is the y-value of the vertex.

Vertex Form of Quadratic Function

Interestingly, we can rewrite a quadratic function using the vertex and the leading coefficient, and the resulting vertex form is f(x) = a(x - h)2 + k. Pay attention to the signs in the vertex form. For example, if the vertex is at (-3,2) and a = 1, the equation would be f(x) = (x + 3)2 + 2. However, if the vertex was at (3,-2), the signs in the equation would change and result in f(x) = (x - 3)2 - 2.

Graph of a Function in Vertex Form
Vertex form and its graph

Factored Form of Quadratic Function

The last form of a quadratic function that can be used to model a real-world scenario is factored form f(x) = a (x - r1)(x - r2), where r1 and r2 are the zeros (x-intercepts) of the function. Remember that the factors need to be set equal to zero and solve for x to 'see' the actual x-intercept.

For example, if f(x) = (x - 2)(x + 3), then x - 2 = 0 and x = 2. Similarly, x + 3 = 0, thus x = -3.

Graph of a Function in Factored Form
Graph of a factored form

You can also graph any of the forms on a graphing calculator and read the key characteristics that are not visible from the equation!

Now that we have refreshed your memory about the quadratic forms and elements, let's use them!

Application of Quadratic Function

Problem 1

You just bought yourself a brand new soccer ball and you want to know how high you can kick it. You know you kicked the ball from the ground level, which is the y-intercept and first zero at the same time (0,0). You estimated that the ball hit the ground approximately 60 feet from where you kicked it (that will be the second x-intercept). Your friend estimated that when the ball was 5 feet away from you, it was about 10 feet up in the air (additional point on the graph, (5,10)).

Now let's figure out the given: zeros: (0,0) and (0,60), and another point (5,10).

Form to use: factored form y = a(x - r1)(x - r2)

Substitute the zeros for the r1 and r2 and the other point for x and y.

Unknown: the vertex and the a value

So we are going to have to solve for a in order to find the vertex.

Solution Explanation
10 = a(5 - 0)(5 - 60) Plug in the numbers: (5,10) for x and y, and the x-intercepts for the r values
10 = a(5)(-55) Evaluate the parentheses first
10 = -275a Multiply
10 / (-275) = a Divide both sides by -275
a = -.036 Simplify

As expected, the leading coefficient is negative, the ball will reach the maximum and hit the ground (obviously!) following a path of a parabola that opens down.

So now we can use the a value and the zeros to rewrite the factored form f(x) = -.036(x - 0)(x - 60) of the trajectory of the ball you kicked. We do not see the actual vertex in the factored form, but no fear, we have technology to help us out! Use any graphic calculator, handheld or online, to graph f(x) and find the highest y value either in the table or on a graph. Here is what you should see. The maximum height of the kick was approximately 32.4 feet.

Graph of the Kicked Ball
Graph of the kicked ball

Problem 2

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