Outcome: C1  model realworld phenomena using quadratic functions
Originator: Kristen
Explanation: Quadratic functions produce parabolas. A parabola is the set of points that are the same distance away from a line (the directrix) and a point (the focus).
Many things in the real world can be modeled using quadratic functions. For example, throwing a ball into the air follows a parabolic path. This path could be represented by a quadratic function that has a negative stretch. A negative stretch would produce a parabola with a vertex being the maximum point. This would make sense for throwing a ball because the ball goes into the air, reaches a maximum height and then comes back down.
Parabolas can also be used to focus light. Flashlights and car headlights have parabola shaped mirrors behind them. Each beam of light that hits the parabola is reflected into the same focus point (as in the graphic above). This allows the light to be focused into one beam, instead of spreading out everywhere and being less effective. It is said that Archimedes used this technique to burn enemy ships before they came to shore. He built a parabola of mirrors and used it to focus light on the ship, eventually causing it to catch on fire. This same technique has been used to cook and burn things. For more information, see the links at the bottom of the page.
Sample Problem:
Sarah throws a softball into the air. The relationship between the softball’s height in metres (y) and time in seconds (x) is represented by the quadratic function y=4x^{2 }+10x+1.5.
a) What is the maximum height that the softball reaches?
b) At what time does the softball hit the ground?
c) At what height will the ball be after 2 seconds?
Solution:
a) The maximum height of the ball will be equal to the vertex of the parabola, since this equation has a negative stretch. To find the x coordinate of the vertex, we can use the formula h=b
2a
y=ax^{2}+bx+c
y=4x^{2}+10x+1.5
0=4x^{2}+10x+1.5
h=b
2a
h= 10
2(4)
h=1.25
This means that at x=1.25, or 1.25 seconds, the ball will be at its maximum height. To find the height, we plug the x value into the equation.
y=4x^{2}+10x+1.5
y=4(1.25)^{2}+10(1.25)+1.5
y=7.75
The maximum height the softball reaches is 7.75 metres.
b) To figure out what time the ball hits the ground, we need to figure out the roots of the equation. We can do this by making the equation equal to 0 and then using the quadratic formula.
y=4x^{2}+10x+1.5
0=4x^{2}+10x+1.5
a=4, b=10, c=1.5
x= b±Ö (b^{2}4ac)
2a

x=10±Ö(10^{2}4(4)(1.5))
2(4)

x= 10±Ö(2400)
2(4)



x=4.87 is an inadmissible answer because the ball could not hit the ground at a negative amount of seconds. Therefore the ball hit the ground at 7.37 seconds.
c) To figure out what height the ball will be after 2 seconds, we insert 2 as the x value in the equation and then solve for y.
y=4x^{2}+10x+1.5
y=4(2)^{2}+10(2)+1.5
y=16+20+1.5
y=5.5
After 2 seconds, the ball is at a height of 5.5 metres.
Comments (1)
Anonymous said
at 8:05 pm on Mar 13, 2007
Excellent work Kristen.
You don't have permission to comment on this page.