<p>There’s a football game in which a player throws a ball from the end zone. It travels vertically going from 3 feet to 10 feet to 50 feet. The vertical distance from the end zone for these three values is 0 feet to 2 feet to 14 feet. Find the equation of the line if the vertical distance varies quadratically with the horizontal distance.</p>
<p>So from that I gathered that the three points were (0, 3) (2, 10) and (14, 50). From there I don’t know what to do.</p>
<p>Another part of the question is that there is a 6 foot fence 295 feet from where the player through it from. Will it go over the fence? And is it possible for someone to catch it before? Where will it finally land?</p>
<p>ANY help is appreciated.</p>
<p>If I am understanding you correctly,</p>
<p>The equation of a parabola is y = ax^2 + bx + c
Plug in the three points and get three equations:
3 = 0a + 0b + c
10 = 4a + 2b + c
50 = (14^2)a + 14b + c</p>
<p>Solve this system of equations:
4a + 2b = 7
196a + 14b = 47</p>
<p>You will find that the equation of your parabola is y = (-1/84)x^2 + (74/21)x + 3</p>
<p>As for the fence, plug in 295 for x. If y is greater than 6 then yes, it will go over the fence. To find where it will finally land, plug in 0 for y.</p>
<p>Ok, thank you very much. I found that it will be at approx. 6.5 feet when it is thrown 295 feet. I am not familiar with football, so it is possible to catch a ball at that height? I’d imagine that it is.</p>
<p>It’s hard to say. The question is worded poorly.</p>
<p>It seems that it will go over the fence and someone may be able to catch it.</p>