Graph the equation. #(3x^2)-(2y^2)-9x+4y-8=0# What are all applicable points (vertex, focus, center, etc)?

The equation can be reorganized to the standard form

Center C(1.5, 1)

Major axis y = 1

Transverse axis x = 1.5

Semi major axis a = sqrt(17/4)=2.062

Vertices on major axis, distant a from C:

Semi transverse axis b = sqrt(51/8) = 2.525

Eccentricity e = sqrt(1+b^2/a^2) = 1.581,

Foci on the major axis, distant ae from C#

Foci S(4.760, 1) and S'(-1.760, 1)

Directrices parallel to transverse axis, distant a/e from it:

Directrrices x = 2.804 and x = 0.196

graph{3x^2-2y^2-9x+4y-8=0 [-10, 10, -5, 5]}

The given equation represents a hyperbola. To graph it and find its properties, we need to rewrite it in standard form. The standard form of the equation of a hyperbola is:

[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 ]

Where ((h,k)) is the center of the hyperbola, (a) is the distance from the center to either vertex along the x-axis, and (b) is the distance from the center to either vertex along the y-axis.

After rearranging the given equation, we get:

[ \frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1 ]

Comparing this to the standard form, we find that:

[ h = 3 ] [ k = 2 ] [ a^2 = 9 ] [ b^2 = 4 ]

This gives us the center at ((3, 2)), (a = 3), and (b = 2).

Now, we can graph the hyperbola, with the center at ((3, 2)), and the vertices at ((3 \pm a, 2)), which are ((6, 2)) and ((0, 2)).

The foci of the hyperbola are given by (c = \sqrt{a^2 + b^2}), so (c = \sqrt{9 + 4} = \sqrt{13}). Therefore, the foci are at ((3 \pm \sqrt{13}, 2)), which are approximately ((6.61, 2)) and ((-0.61, 2)).

In summary, the applicable points are:

• Center: ((3, 2))
• Vertices: ((6, 2)) and ((0, 2))
• Foci: Approximately ((6.61, 2)) and ((-0.61, 2))