Question 1006121
9x^2 -4y^2 +36x -16y -16 = 0
this is the equation of a hyperbola
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factor equation and complete the square
9(x+2)^2 -4(y+2)^2 = 36
now divide both sides of = by 36
(x+2)^2 / 4 - (y+2) / 9 = 1
this is standard form for hyperbola
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center of hyperbola is (-2, -2)
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vertices are (h+a,k) and (h-a,k), (0, -2) and (-4, -2)
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to determine foci, first calculate c^2 = a^2 + b^2
c^2 = 4 + 9
c = sqrt(13)
foci are (-2+sqrt(13), -2) and (-2-sqrt(13))
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slope of asymptotes are + or - b/a = 9/4 and -9/4
equations of asymptotes are y = k + or - b/a(x-h),
y = -2 + (9/4)(x - 2) and y = -2 - (9/4)(x - 2)
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here is the graph
{{{ graph(300, 200, -6, 2, -6, 2, (1/2)(-3sqrt(x^2+4x)-4), (1/2)(3sqrt(x^2+4x)-4), -2+(9/4)(x+2), -2-(9/4)(x+2))}}}
note that the left curve of the hyperbola is continuous, not sure why it is broken