Question 695436
Hyperbola:
9x^2-16y^2-18x-32y-151=0
9x^2-18x-16y^2-32y-151=0
complete the square: (you already did with a small error which I have corrected)
This is a hyperbola with horizontal transverse axis: (x-term listed first)
Its standard form of equation: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center.
Center:(1,-1)
a^2=16
a=√16=4
Vertices: (1±a,-1)=(1±4,-1)=(-3,-1) and (5-1)
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b^2=9
b=√9=3
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c^2=a^2+b^2=16+9=25
c=√25=5
Foci: (1±c,-1)=(1±5,-1)=(-4,-1) and (6-1)
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Asymptotes:(straight line equations that go thru center, of the form y=mx+b, m=slope, b=y-intercept)
For hyperbolas with horizontal transverse axis: slope=±b/a=±3/4

Equation for asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates of center.
-1=-3*1/4+b
b=-1/4
equation:y=-3x/4-1/4
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Equation for asymptote with positive slope:
y=3x/4+b
solve for b using coordinates of center.
-1=3*1/4+b
b=-7/4
equation:y=3x/4-5/4
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Graph: 
9(x^2-2x+1)-16(y^2+2y+1)=151-16+9 
9(x-1)^2-16(y+1)^2=144
(x-1)^2/16-(y+1)^2/9=1
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See graph below:
y=±((9(x-1)^2/16)-9)^.5-1
{{{ graph( 300, 300, -10, 10, -10, 10,((9(x-1)^2/16)-9)^.5-1,-((9(x-1)^2/16)-9)^.5-1,-3x/4-1/4,3x/4-5/4) }}}