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anonymous

  • one year ago

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?

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  1. anonymous
    • one year ago
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    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

  2. campbell_st
    • one year ago
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    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.

  3. anonymous
    • one year ago
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    okay, I'm getting it so far

  4. campbell_st
    • one year ago
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    |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

  5. anonymous
    • one year ago
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    so the focal length would be 4?

  6. campbell_st
    • one year ago
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    |dw:1436904321116:dw|

  7. campbell_st
    • one year ago
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    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..?

  8. anonymous
    • one year ago
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    (8, -7) ?

  9. campbell_st
    • one year ago
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    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

  10. campbell_st
    • one year ago
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    which would require you to divide both sides on the equation by 4 then subtract 7 from both sides...

  11. anonymous
    • one year ago
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    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

  12. campbell_st
    • one year ago
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    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)

  13. anonymous
    • one year ago
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    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 \[f(x) = \frac{ 1 }{ 4 } (x-1)^{2} +1\]

  14. anonymous
    • one year ago
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    so would it be the first one?

  15. anonymous
    • one year ago
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    I'm sorry I'm a moron i was looking at the wrong question....

  16. anonymous
    • one year ago
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    but would you mind walking me through the other two that i posted?

  17. campbell_st
    • one year ago
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    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...

  18. anonymous
    • one year ago
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    yeah i was looking at my options for the third question not the first...

  19. campbell_st
    • one year ago
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    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...

  20. anonymous
    • one year ago
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    is the equation the same no matter which way the parabola is facing?

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