Answer:
hyperbola with a solid line boundary
(4, -2)
(0, -4)
Step-by-step explanation:
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The shape of the given equation is hyperbola with solid line boundary, key feature i.e. the center is at (4, -2), and ordered pair is (0, -4) and the standard form will be [tex]\frac{(x-4)^2}{9} -\frac{(y+2)^2}{4} \leq1[/tex] .
Conic section is a curve obtained by the intersection of the surface of a cone with a plane.
we have,
[tex]4x^2-32x-9y^2-36y-8\leq 0[/tex]
And,
The general form of a Conic Section is,
[tex]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/tex]
And,
The Standard form of a is,
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1[/tex]
So,
Now,
[tex]4x^2-32x-9y^2-36y-8\leq 0[/tex]
Then,
[tex]4x^2-32x-9y^2-36y\leq8[/tex]
Now, taking common,
[tex]4(x^2-8x)-9(y^2+4y)\leq8[/tex]
Now,
Adding 64 and subtracting 36 to both sides ,
We get,
[tex]4(x^2-8x+16)-9(y^2+4y+4)\leq8+64-36[/tex]
Now,
factor the perfect square,
i.e.
[tex]x^2-8x+16=(x-4)^2[/tex]
Now,
[tex]y^2+4y+4=(y+2)^2[/tex]
Now,
[tex]4(x-4)^2-9(y+2)^2\leq36[/tex]
Now,
Divide both sides by 36,
i.e.
[tex]\frac{4(x-4)^2}{36} -\frac{9(y+2)^2}{36} \leq\frac{36}{36}[/tex]
We get,
[tex]\frac{(x-4)^2}{9} -\frac{(y+2)^2}{4} \leq1[/tex]
So, this is the standard form of the given cubic inequality.
Hence, we can say that the shape of the given equation is hyperbola with solid line boundary, key feature i.e. the center is at (4, -2), and ordered pair is (0, -4) and the standard form will be [tex]\frac{(x-4)^2}{9} -\frac{(y+2)^2}{4} \leq1[/tex] .
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