Please help algebra 2
Tony is trying to find the equation of a quadratic that has a focus of (−1, 4) and a directrix of y = 8. Describe to Tony your preferred method for deriving the equation. Make sure you use Tony's situation as a model to help him understand.
first start with the vertex-form of a parabola, its: y = a(x-h)^2 + k next, solve for the vertex of the quadratic by using the value of the focus and the directrix. since, the focus is: (-1, 4) and the direxrix is: y=8 use the mid point formula from the focus to the directrix to find the y-value of the vertex. the x-value of the vertex will be the same as the x-value in the focus, -1 so at this point, we will have the values of the vertex, (h, k). plug those into the equation. all that is left to find is the value for 'a'
Would the midpoint be -0.5/ 2?
@DemolisionWolf
hmm.... maybe it would be easier if I showed it this way: |dw:1387221229918:dw| where is the midpoint for this number line, given the two points, i circle?
the midpoint would be 6
@DemolisionWolf
yep!
so we know that the vertex of the parabola is (h, k) or (-1, 6) plug those into the equation: y = a(x-h)^2 + k
y = a(-1 - 6)^2 + k?
close!! (h, k) is (-1, 6) so, h = -1 k = 6 try again ^_^ once more
y = a(x- (-1))^2 + 6
yep, thats right
so, two more steps, the next step is to figure out what 'a' is. then once we have that, we expand the equation we have been 'making'
Could you keep explaining how to solve? I know I may be frustrating you but I really do want a good grade in my class.
yea, i can do that
do you have any question about why i've done the steps I have so far?
Nope, I think I've gotten it. I'm writing it down on a sheet of paper and following along :)
@DemolisionWolf
cool
so the next step was to find 'a' 'a' is the distance from the vertex (-1, 6) to the focus (-1, 4) can you tell me what that distance is?
12?
@DemolisionWolf
|dw:1387222999345:dw| try again ^_^
Would it be -1, 5? or just 5? I used the distance formual to get 12 lol
@DemolisionWolf
did u get my message i sent? a=-2
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