anonymous
  • anonymous
how would i create a parabola touching end-to-end inside a rectangle that is 14'' long and 5'' high
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1377981925206:dw|
anonymous
  • anonymous
kinda like that
nincompoop
  • nincompoop
you just did it

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anonymous
  • anonymous
that was just a drawing. how would i find the coordinates?
anonymous
  • anonymous
or the equation?
nincompoop
  • nincompoop
do you know where the focus point is?
anonymous
  • anonymous
no i have no clue
anonymous
  • anonymous
(0,5) vertex ((7,0), and othe side (14, 5)
nincompoop
  • nincompoop
oh so you have the coordinates!
nincompoop
  • nincompoop
the standard form of a quadratic function is: \[f(x)=a(x-h)^2+k; a \neq0\] The graph of f is a parabola whose axis is the vertical line x = h and whose vertex is the point (h, k). If a > 0, the parabola opens upward, and if a < 0, the parabola opens downward
anonymous
  • anonymous
would this be correct? i think this is it \[y=.08(x-7)^2\]
jdoe0001
  • jdoe0001
|dw:1377984257262:dw| so we could say that we know 2 points of it, (-7, 5) and (7, 5) if we use the "vertex form" of a parabola, that is \(\bf y = a(x-h)^2 + k\) we could use either of those points and the origin as our vertex to find what "a" component is, that is \(\bf y = a(x-h)^2 + k\\ \textit{say using (-7, 5)}\\ 5 = a(-7-0)^2+0\)
jdoe0001
  • jdoe0001
assuming you weren't given any specific points per se
anonymous
  • anonymous
oh ok, so now that i looked back at my work it wasnt a height of 5 but rather 4
anonymous
  • anonymous
thats why i thought it was y=.08(x-7)^2
anonymous
  • anonymous
but lets say that is the correct equation, how would i find the focus?
jdoe0001
  • jdoe0001
hmmm, you use the "focus form" of it :S \(\bf (x-h)^2=4p(y-k)\) and one can say that \(\bf y=.08(x-7)^2 \implies \cfrac{1}{0.08}(y-0) = (x-7)^2\)
jdoe0001
  • jdoe0001
so that would make \(\bf 4P = \cfrac{1}{0.08}\) P = distance from the vertex to the focus
anonymous
  • anonymous
thank you!!!1

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