A football field appears to be flat, its surface is actually shaped like a parabola so that rain runs off to both sides. The cross section of a field with synthetic turf can be modeled by y = -0.000234x(x - 160) where x and y are measured in feet.
a) What is the fields length?
b) What is the maximum height of the field's surface? @whpalmer4

Question

A football field appears to be flat, its surface is actually shaped like a parabola so that rain runs off to both sides.  The cross section of a field with synthetic turf can be modeled by y = -0.000234x(x - 160) where x and y are measured in feet.
a) What is the fields length?
b) What is the maximum height of the field's surface? @whpalmer4

whpalmer4 · Answer

$$y = -0.000234x(x-160)$$I guess the x axis runs the length of the field, so one end is at x = 0, and the other end is wherever y = 0 (other than at x =0).  Do you agree?

anonymous · Answer

@whpalmer4 yes

anonymous · Answer

@whpalmer4 well ok so this question comes with a graph do you want me to upload it?

anonymous · Answer

attachment

anonymous · Answer

@whpalmer4 i don't know if it helps you

whpalmer4 · Answer

Yes, that's what I was envisioning.  

Okay, so how do we find the field's length?

whpalmer4 · Answer

The length of the field is the distance between the two points where the curve crosses the x-axis, right?

whpalmer4 · Answer

If we plug in $x = 0$, we have $$y = -0.000234(0)(0-160) = 0$$

anonymous · Answer

@whpalmer4 ok so how would i find the maximum height of the field's surface?

whpalmer4 · Answer

Let's expand the formula:

$$y = -0.000234x^2 + 0.03744x$$$$0 = -0.000234x^2+0.03744x$$We already found $x=0$ as one solution, but we need the other.  Probably the easiest thing here is to use the quadratic formula:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$which gives the solutions to $$ax^2+bx+c=0,\,a\ne0$$
We have $$a = -0.000234,b = 0.03744,c=0$$
$$x = \frac{-0.03744\pm\sqrt{0.03744^2-4(-0.000234)(0)}}{2(-0.000234)}$$which simplifies to $$x=\frac{-0.03744\pm 0.03744}{-0.000468}$$

whpalmer4 · Answer

I think you can get the non-zero value of $x$ from that, right?

Now to find the maximum height, which is at the vertex of the parabola.  There are a number of ways we could do this.  One is to remember that a parabola is symmetric about its vertex, and so the vertex will be centered between the two values of $x$ that make $y=0$.  Divide the length of the field by 2, in other words, and that's the x-coordinate of the vertex.  Plug that value of $x$ into the formula for the height of the cross-section, and you've got your maximum height.

whpalmer4 · Answer

Now, if you don't remember that the vertex is centered between the two x-intercepts, you might remember that the x-coordinate of the vertex where the equation is written in the form $y = ax^2 + bx + c$ is at $$x = -\frac{b}{2a}$$which would give us an x-value of $$x=-\frac{0.03744}{-0.000468} = $$

yet another way would be to use the "vertex form" of a parabola, which is 
$$y = a(x-h)^2+k$$with the vertex at $(h,k)$Putting our equation in comparison:
$$y=a(x-h)^2+k$$$$ y = -0.000234x^2 + 0.03744x$$we would have to complete the square to make that work.  Let's not, because we'd just get the same answer for a bunch of extra work :-)

whpalmer4 · Answer

Actually, looking at the problem, we could have done this more easily.  

$$y = -0.000234x(x-160)$$The solutions aren't affected at all by the -0.000234 out front: that only affects how high the parabola goes (or how low), but not where it crosses the x-axis!

Really, we just need to find the solutions to $$0 = x(x-160)$$And that's easy!$$0=x(x-160)$$is solved by $$x=0$$$$x-160 = 0$$so our solutions are $$x = 0, 160$$

Half the length of the field is the midpoint, at x = 80.  The height there is 
$$-0.000234(80)(80-160) = -0.000234*-6400 = 1.4976$$

whpalmer4 · Answer

My eyes just glazed over when I saw the big ugly decimal out front, and kept me from spotting the easier path to the solution :-)

anonymous · Answer

@whpalmer4 I'm sorry you had to go through all of this

anonymous · Answer

@whpalmer4 Thank you you're the best :D

whpalmer4 · Answer

Considering what you had for the prior two problems, I should have tried to see it in the same fashion.

No problem, a little extra work is good for me :-)

whpalmer4 · Answer

Did I succeed in thoroughly confusing you?

anonymous · Answer

@whpalmer4 Yes you did i was lost :O i was going to ask you if there was another way you could of help me because i was in confusion haha thank you so much :D

whpalmer4 · Answer

Let me try to make it as clear as possible, now that I've made it as unclear as possible :-)

$$y = -0.000234x(x-160)$$ is the height of the field at any point along its length.  The ends of the field are the two points where the height is 0.
$$y=-0.000234x(x-160)$$is 0 when either $$-0.000234x = 0$$which happens at $x =0$or when $$(x-160) = 0$$which happens when $$x = 160$$

So, the length of the field is the distance between x = 0 and x = 160, which is 160.  Clear?

anonymous · Answer

@whpalmer4 ok so a) is 160 and b) is 1.4976 right?

whpalmer4 · Answer

Now, the formula is that of a parabola, which has its vertex equidistant between the two x-intercepts.  That means the vertex has an x-coordinate of $$x=\frac{0+160}{2} = 80$$Now we just find the height of the field at $x=80$ and we are all done:
$$y = -0.000234(80)(80-160) = 1.4976$$

What we are actually looking at is the distance across the football field (from side to side), not the length of the football field.  Problem is written a bit misleadingly, in my opinion.  At first I was wondering what kind of an idiot would think a football field is only 160 feet long :-)

whpalmer4 · Answer

Yes, a is 160, b is 1.4976.  Probably 1.5 is close enough, considering it's built on a pile of dirt :-)

whpalmer4 · Answer

I hope that cleared it up, I've got to go!

anonymous · Answer

@whpalmer4 Thank you so much you didn't have to explain all of this thank you for taking the time to bare with me i appreciate it :D <3 :*