Question

Help me please!!?!?? ?

Answer 1

1

$3x^2 + 5y^2 - 12x + 30y + 42 = 0$

$3x^2 - 12x + 5y^2 + 30y = - 42$

$3(x^2 - 4x) + 5(y^2 + 6y) = -42$

$3(x^2 -4x + 4) + 5(y^2 + 6y + 9) = -42 + 12 + 45$

$3(x-2)^2 + 5(y+3)^2 = 15$

$\dfrac{(x-2)^2}{5} + \dfrac{(y+3)^2}{3} = 1$

Ellipse, same signs different coefficients on $x^2$ and $y^2$

2

$9x^2 -36x - 4y^2 + 8y = 4$

$9(x^2 - 4x + 4) - 4(y^2 - 2y + 1) = 4 + 36 - 4$

$9(x- 2)^2 - 4(y - 1^2) = 36$

Hyperbola, opposite signs on $x^2$ and $y^2$

3

$y = x^2 +2x + 3 = (x + 1)^2 + 2$

parabola

4

$x^2 - 4 x + y^2 + 4y = 4$

$(x - 2)^2 + (y+2)^2 = 4 + 4 + 4 = 12$

equal coefficents on $x^2$ and $y^2$, circle

5

$-9x^2 - 18x + 4y^2 - 8y = 41$

$-9(x^2 -2x + 1) + 4(y^2 - 2y + 1) = 41 - 9 + 4$

$-9(x-1)^2 + 4(y-1)^2 = 36$

hyperbola, opposite signs

6

$-4x = y^2 - 2y - 11 = (y-1)^2 - 12$

That's a parabola, sideways open to the left

7

$2(x^2 + 6x) + 3(y^2 - 8y) = -60$

$2(x^2 + 6x + 9) + 3(y^2 - 8y + 16) = -60 + 18 + 48$

$2(x+3)^2 + 3(y - 4)^2 = 6$

$\dfrac{(x+3)^2}{3} + \dfrac{(y-4)^2}{2} = 1$

Ellipse, different coefficients

8

$16x^2 - 32x - y^2 - 6y = 57$

$16(x^2 - 2x + 1) - (y^2 + 6y + 9) = 57 + 16 - 9 = 64$

opposite signs, hyperbola

You'll have to do the graphing yourself, sorry