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Mathematics 11 Online
OpenStudy (anonymous):

omg can someone please just HELP ME!

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

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 :(

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

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

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?

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 ???

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

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

OpenStudy (anonymous):

oooh yeaaah! thanksss

jimthompson5910 (jim_thompson5910):

you're welcome

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