anonymous
  • anonymous
What is the equation of the quadratic graph with a focus of (8, −8) and a directrix of y = −6? and can someone show me how to solve problems like this?
Mathematics
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anonymous
  • anonymous
What is the equation of the quadratic graph with a focus of (8, −8) and a directrix of y = −6? and can someone show me how to solve problems like this?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
i have two others like this What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6? and What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1? but if someone could just tell me how to solve these that'd be cool too
campbell_st
  • campbell_st
ok... there are several ways to do this I start by graphing things so I know what I'm looking at |dw:1436903991982:dw| so looking at things... the directrix is above the focus so the parabola is concave down. so then I know the general form of the curve is \[(x -h)^2 = -4a(y - k)\] where (h, k) is the vertex and a is the focal length.
anonymous
  • anonymous
okay, I'm getting it so far

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campbell_st
  • campbell_st
|dw:1436904182134:dw| What is the distance between the focus y = -6 and the y value in the focus y = -8 this is double to focal length or 2a... so if you can find it, then halve the answer you have the focal length
anonymous
  • anonymous
so the focal length would be 4?
campbell_st
  • campbell_st
|dw:1436904321116:dw|
campbell_st
  • campbell_st
close its y = -6 to y = -8 so the distance is 2 units so 2a = 2 or a = 1 so the focal length a = 1 next the vertex is 1 unit above the focus on the line of symmetry vertex is at (8, -8 + a) where do you think the vertex is..?
anonymous
  • anonymous
(8, -7) ?
campbell_st
  • campbell_st
great so substituting the vertex and focal length into the general form you would have \[(x - 8)^2 = 4 \times 1 \times (y +7)\] or \[(x -8)^2 = 4(y + 7)\] you may have to rewrite the equation with y as the subject
campbell_st
  • campbell_st
which would require you to divide both sides on the equation by 4 then subtract 7 from both sides...
anonymous
  • anonymous
okay that lost me a little bit... my answer starts with f(x) as the subject which I'm pretty sure is the same as y but none of my options match what i got
campbell_st
  • campbell_st
ok if you have \[(x-8)^2 = 4(y + 7)\] divide both sides by 4 \[y + 7 = \frac{1}{4}(x -8)^2 \] now subtract 7 from both sides and change y to f(x)
anonymous
  • anonymous
so i got \[f(x) = \frac{ 1 }{ 4 } (x-8)^{2} + 7\] but my only options are \[f(x) = \frac{ 1 }{ 4 } (x-1)^{2}\] and \[f(x) = \frac{ 1 }{ 4 } (x-1)^{2} +1\]
anonymous
  • anonymous
so would it be the first one?
anonymous
  • anonymous
I'm sorry I'm a moron i was looking at the wrong question....
anonymous
  • anonymous
but would you mind walking me through the other two that i posted?
campbell_st
  • campbell_st
oops I forgot the negative it should be \[f(x) = -\frac{1}{4}(x -8)^2 - 7\] as the parabola is concave down. I think the answer choices you posted don't match the question you posted...
anonymous
  • anonymous
yeah i was looking at my options for the third question not the first...
campbell_st
  • campbell_st
so there is a solution for your question.. if you need help with others in the same form, you'll need to post them as seperate questions... always start by plotting the information you are given, that way you'll know the concavity of the parabola. then work from there...
anonymous
  • anonymous
is the equation the same no matter which way the parabola is facing?

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