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Question 572346: Please help me solve and graph this equation: 
I tried putting it in standard form and it ended up being a horizontal hyperbola. The standard form was 
From there I got the center of the hyperbola which is (h,k) or (-2,-4)and I graphed it. Then I looked for the values for a, b, and c. A was  or 2.6 as a decimal and B was  or 7.9 as a decimal. I graphed A and B as the conjugate and transverse axes. This is where I got confused. The equation for the hyperbola had x on the left and y on the right, meaning it should graph horizontally. But I graphed the axes and the hyperbola came out vertical. Please help me see where I went wrong!
Found 2 solutions by lwsshak3, KMST:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve and graph
9x^2-4y^2+18x+32y-91=0
complete the square
9(x^2+2x+1)-4(y^2-8y+16)=91+9-64=36
9(x+1)^2-4(y-4)^2=36
(x+1)^2/4-(y-4)^2/9=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
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For given equation:
Center:(-1,4)
a^2=4
a=2
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b^2=9
b=3
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slope of asymptotes=±b/a=±3/2
Equation of asymptotes:
y=mx+b
y=-3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=-(3*-1)/2+b
b=5/2
Equation:y=-3x/2+5/2
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y=3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=(3*-1)/2+b
b=11/2
Equation:y=3x/2+11/2
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see the graph below as a visual check on the above:
y=(2.25(x+1)^2-9)^.5+4
Answer by KMST(5328)  (Show Source):
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