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I have to graph this parabola. (5y^2)+10y-x+4=0 Can somebody show me the steps on how to do this
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are you sure its entered coorectly?
yes.
then its most likely e geometric equation of a parabola
set it equal to x x = 5y^2 +10y +4 this will give us a sideways parabola
complte the square for the ys
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right do you know how to find x and y intercepts?
is it this ( -b/2a , 4ac-b^2/4a) ?
y^2 +2y = x/5 -4/5
find vertex using y = -b/2a where a,b,c are the coefficients
(y+1)^2 = x/5 - 4/5 +1
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... x/5 + 1/5 =1/5(x+1)
(y+1)^2 = (1/5)(x+1) has the form y^2 = 4ax 1/5 = 4a a = 1/20
the center is at (-1,-1) ang opens to the right
by center I mean vertex ;)
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