The Quadratic Triple Axel

I hope I don’t fuddle you đŸ™‚

I did this for school.

How are the real solutions of a quadratic equation related to the graph of the quadratic function?

For those of you who don’t understand ice skating, I have provided explanations.

In ice skating, there is an advanced jump called triple Axel. When I do any jump, I can model the parabola that my feet make in the air with a quadratic equation. In the air, I rotate with my left foot crossed over my right, and my legs are straight. My feet make a certain path starting from the point where I “take off” to when I “land”. (In skating language, “takeoff” is the moment right when you jump. “Landing” is when your feet touch the ice again.)

The minimum jump height required for a triple axel is 18 inches. Of course, I have to have a certain speed going into it, which will make my triple axel travel a certain distance over the ice. Say I did a triple axel where I jumped 18 inches high and I got 48 inches of distance. I can write a quadratic function to model this.

I reach my maximum height halfway through the jump, which means that I am 18 inches off the ice when I have traveled 24 inches in the air.

In vertex form, which is y = a(x – h)^2 + k, h is 24 and k is 18.

y = a(x – 24)^2 + 18

(0,0) is a point on the graph of the path of my supposed triple axel. (0,0) is the takeoff point. (48,0) is the landing point.

Substitute (0,0) into the currently unfinished vertex form function to find the vertical stretch or compression factor a.

0 = a(0 – 24)^2 + 18

0 = a(-24)^2 + 18

0 = 576a + 18

576a = -18

a = -1/32

a = -0.03125

The quadratic function for this is y = -0.03125(x – 24)^2 + 18.

The parabola opens downwards.

The vertex is (24, 18). The maximum value is 18 and the axis of symmetry is x = 24.

The maximum value is 18.

The domain can be modeled by:

x is greater than or equal to 0 and less than or equal to 48.

The range can be modeled by:

y is greater than or equal to 0 and less than or equal to 18.

The two solutions are (0,0), (48,0), and the vertex, which is a maximum value, is at (24,18). This means that right at my takeoff, I have jumped 0 inches. When my jump distance has reached 48 inches, I have landed. When my distance is 24 inches, I have reached my maximum height, 18 inches.

I graphed the jump path of my imaginary triple Axel on a Cartesian plane. I hope I can actually do one like that on the ice someday!

If all of that is too much of a headache for you, here’s something funny!

I am doing an ice bath for my foot. Snooki always wants to join me. I keep on trying to tell her that it’s not fun.

By Vincent Zhou

One comment on “The Quadratic Triple Axel

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s