Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

I have to graph this parabola. (5y^2)+10y-x+4=0 Can somebody show me the steps on how to do this

OpenStudy (amistre64):

are you sure its entered coorectly?

OpenStudy (anonymous):

yes.

OpenStudy (amistre64):

then its most likely e geometric equation of a parabola

OpenStudy (dumbcow):

set it equal to x x = 5y^2 +10y +4 this will give us a sideways parabola

OpenStudy (amistre64):

complte the square for the ys

myininaya (myininaya):

right do you know how to find x and y intercepts?

OpenStudy (anonymous):

is it this ( -b/2a , 4ac-b^2/4a) ?

OpenStudy (amistre64):

y^2 +2y = x/5 -4/5

OpenStudy (dumbcow):

find vertex using y = -b/2a where a,b,c are the coefficients

OpenStudy (amistre64):

(y+1)^2 = x/5 - 4/5 +1

OpenStudy (amistre64):

... x/5 + 1/5 =1/5(x+1)

OpenStudy (amistre64):

(y+1)^2 = (1/5)(x+1) has the form y^2 = 4ax 1/5 = 4a a = 1/20

OpenStudy (amistre64):

the center is at (-1,-1) ang opens to the right

OpenStudy (amistre64):

by center I mean vertex ;)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!