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OpenStudy (anonymous):
using a directrix of y = -3 and a focus of (2,1) what quadratic function is created
jimthompson5910 (jim_thompson5910):
y = -3 is 4 units away from y = 1 (the y coordinate of the focus) since |-3 - 1| = |-4| = 4
Cut this in half to get 4/2 = 2
Therefore the focal distance is p = 2. This means that the vertex is 2 units away from the focus. So we subtract 2 from the y coordinate of y = 1 to get 1-2 = -1
The vertex is (2,-1) ----> h = 2, k = -1
Now plug all this into the equation below
4p(y-k) = (x-h)^2
4(2)(y-(-1)) = (x-2)^2
8(y+1) = (x-2)^2
jimthompson5910 (jim_thompson5910):
Optionally you can solve for y to get
8(y+1) = (x-2)^2
y+1 = (1/8)(x-2)^2
y = (1/8)(x-2)^2 - 1
OpenStudy (anonymous):
Thank you so much..can you help with another one!
OpenStudy (anonymous):
simplify y^2+7y +2/y^2-2y-15
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jimthompson5910 (jim_thompson5910):
is there a typo in the numerator?
OpenStudy (anonymous):
oops yes
OpenStudy (anonymous):
y^2 + 7y + 12/y^2-2y-15
jimthompson5910 (jim_thompson5910):
are you able to factor the numerator?
OpenStudy (anonymous):
idk :(
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jimthompson5910 (jim_thompson5910):
find two numbers that multiply to 12 and add to 7
OpenStudy (anonymous):
4 and 3
jimthompson5910 (jim_thompson5910):
so that means y^2 + 7y + 12 factors to (y+4)(y+3)
jimthompson5910 (jim_thompson5910):
what does y^2-2y-15 factor to?
OpenStudy (anonymous):
hmm thats tricky
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OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
find two numbers that multiply to -15 and add to -2
OpenStudy (anonymous):
ahh ok I got it !!
jimthompson5910 (jim_thompson5910):
so what does it factor to?
OpenStudy (anonymous):
y+4/y-5
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OpenStudy (anonymous):
hey what is it called for htis
OpenStudy (anonymous):
jimthompson5910 (jim_thompson5910):
This is the multiplication of two rational expressions
jimthompson5910 (jim_thompson5910):
First multiply straight across. Then reduce the fraction as much as possible
OpenStudy (anonymous):
what about the diff variables?
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jimthompson5910 (jim_thompson5910):
you would leave them be
in the numerator, you would multiply the coefficients 15 and 6 to get 90
in the denominator you'll have 20*5 = 100
jimthompson5910 (jim_thompson5910):
so far, you have
90x^6y^2
------------
100x^4y^5
jimthompson5910 (jim_thompson5910):
Now reduce
OpenStudy (anonymous):
um...im lost. ..I can only reduce if the bigger num is on top
jimthompson5910 (jim_thompson5910):
90/100 reduces to ???
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OpenStudy (anonymous):
9 and 10
jimthompson5910 (jim_thompson5910):
9/10, yep
jimthompson5910 (jim_thompson5910):
x^6/x^4 reduces to???
OpenStudy (anonymous):
3 and 2!!
jimthompson5910 (jim_thompson5910):
subtract the exponents
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jimthompson5910 (jim_thompson5910):
6-4 = 2
so, x^6/x^4 = x^2
jimthompson5910 (jim_thompson5910):
for the y terms, 5-2 = 3
because the bigger term is in the bottom, we'll have y^3 in the bottom
jimthompson5910 (jim_thompson5910):
that means the final answer is
9x^2
----------
10y^3