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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1ok... 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
 one year ago
Best ResponseYou've already chosen the best response.0okay, I'm getting it so far

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1dw: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
 one year ago
Best ResponseYou've already chosen the best response.0so the focal length would be 4?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436904321116:dw

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1close 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..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1great 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
 one year ago
Best ResponseYou've already chosen the best response.1which would require you to divide both sides on the equation by 4 then subtract 7 from both sides...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.1ok if you have \[(x8)^2 = 4(y + 7)\] 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
 one year ago
Best ResponseYou've already chosen the best response.0so i got \[f(x) = \frac{ 1 }{ 4 } (x8)^{2} + 7\] but my only options are \[f(x) = \frac{ 1 }{ 4 } (x1)^{2}\] and \[f(x) = \frac{ 1 }{ 4 } (x1)^{2} +1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be the first one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry I'm a moron i was looking at the wrong question....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but would you mind walking me through the other two that i posted?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1oops 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah i was looking at my options for the third question not the first...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.0is the equation the same no matter which way the parabola is facing?
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