anonymous
  • anonymous
a bridge is supported by three arches. The function that describes the arches is h(x) = 0.25x^2 + 2.375x where h(x) is the height, in meters, of the arch above the ground at any distance, x, in meters, from one end of the bridge. How far apart are the bases of each arch?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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whpalmer4
  • whpalmer4
\[h(x) = 0.25x^2+2.375x\]The bases are going to be at the points where \(h(x) = 0\). Can you solve \(0 = 0.25x^2+2.375x\)?
whpalmer4
  • whpalmer4
That equation must not be correct. You need a negative term in there somewhere, or it just keeps getting higher....
whpalmer4
  • whpalmer4
Are you sure it isn't \(h(x) = -0.25x^2+2.375x\)?

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anonymous
  • anonymous
yes omg you're right so sorry!
anonymous
  • anonymous
what does that mean!?
whpalmer4
  • whpalmer4
Well, it's going to be an inverted parabola.|dw:1371166791215:dw| at x = 0, that's the end of the bridge. at the other spot where y = 0, that's the first base.
whpalmer4
  • whpalmer4
so, you need to solve the equation to find where h(x) = 0. There are two solutions, although one of them is at x = 0. The other one is the spot where the base goes. Make sense?
whpalmer4
  • whpalmer4
This is also the path that a baseball would follow if you threw it up in the air (with no air resistance, that is)
anonymous
  • anonymous
okay...
anonymous
  • anonymous
whats next
whpalmer4
  • whpalmer4
Well, did you solve the equation?
anonymous
  • anonymous
ahh no, maybe im not getting it-
anonymous
  • anonymous
where is the second base
whpalmer4
  • whpalmer4
Okay, no problem, we'll fix that! :-) So, we want to find the value of \(x\) such that \(h(x) = 0\) to find the other end of our arch. \[h(x) = 0 = -0.25x^2+2.375x\]A little experience tells me that those ugly numbers are all 1/8ths, but even easier is to just multiply through by 1000 and get rid of the decimals altogether. \[0 = -0.25x^2+2.375x\]\[0*1000=-0.25*1000*x^2+1000*2.375x\]\[0=-250x^2+2375x\]We can factor that, or use the quadratic equation, which do you prefer?
anonymous
  • anonymous
im more familiar with factoring, but whichever is best to teach
whpalmer4
  • whpalmer4
Why don't you try factoring that and show me what you get...
anonymous
  • anonymous
would you mind doing it but showing exactly how you got there, its easier for me to see the proper steps, especially with crazy high numbers : $
whpalmer4
  • whpalmer4
Well, do you see any common factors in -250x^2 + 2375x?
anonymous
  • anonymous
5?
whpalmer4
  • whpalmer4
Yes. A whole mess of them, in fact :-) What else?
whpalmer4
  • whpalmer4
As a first step, can't you write x(-250x+2375) ?
anonymous
  • anonymous
25?, 10?
whpalmer4
  • whpalmer4
\[-250x^2+2375x = 0\]Factor out \(x\) \[x(-250x + 2375) = 0\]For that to be true, either \(x=0\) or \(-250x + 2375 = 0\) I bet you can solve both of those equations :-)
anonymous
  • anonymous
ive never factored without a middle number
whpalmer4
  • whpalmer4
\(x=0\) is the case for the bank end of the arch. \(-250x + 2375 = 0\) is the case for the other end of the arch.
whpalmer4
  • whpalmer4
I'm sure you have, you just don't remember it as such. You maybe haven't solved an equation by factoring where that is true, but I guarantee that you must have factored something like \(a^2 + ab\) when you were first learning how to factor :-)
anonymous
  • anonymous
uhhh im so confused, what do i do next ?
whpalmer4
  • whpalmer4
Solve \[-250x + 2375 = 0\]
anonymous
  • anonymous
is it 9.5
anonymous
  • anonymous
i had to use square root factor
whpalmer4
  • whpalmer4
No. No square root needed here. \[-250x + 2375 = 0\]Add 250x to both sides \[-250x + 250x + 2375 = 250x\]\[2375=250x\]Now what do you think is the value of \(x\)?
anonymous
  • anonymous
9.5
whpalmer4
  • whpalmer4
Oh, so sorry...9.5 is correct. I wrote it down as 19/2 :-)
anonymous
  • anonymous
haha thank you soo much!
whpalmer4
  • whpalmer4
I'd say I was testing your confidence, but... :-)
whpalmer4
  • whpalmer4
Okay, so at x = 9.5, the height of the arch should be 0 again. Let's check: -0.25(9.5)^2 +2.375(9.5) = 0 -0.25*90.25+2.375*9.5=0 -22.5625 + 22.5625 = 0 Success!
whpalmer4
  • whpalmer4
Here's a plot of h(x) from x = 0 to x = 10:
1 Attachment
whpalmer4
  • whpalmer4
So, the bases are 9.5 meters apart, because that's how far it is from x = 0 to x = 9.5.
whpalmer4
  • whpalmer4
I'll quickly do it with the quadratic formula, just for illustration: we had \[h(x) = -0.25x^2+2.375x\]so our quadratic would be\[-0.25x^2+2.375x = 0\]giving us \(a=-0.25, b = 2.375, c = 0\). Our solutions would be thus\[x = \frac{-2.375\pm\sqrt{2.375^2-4(0)(-0.25)}}{2*(-0.25)} = \frac{-2.375\pm2.375}{-0.5} = 0, 9.5\]

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