Question

how do you graph (y-2)^2=-2(x+2)

Answer 1

Parabola

Answer 2

What is the foci of this equation? And shouldn't it be opening to the left?

Answer 3

@Directrix

Answer 4

generally for any given equation like (y + h)^2 = (x + k).
Vertex = (-h,-k)

or you could use complete the square for y
\[(y-2)^2=-2 (x+2)\]
\[-2 (x+2)=(y-2)^2\]
\[x=-\frac{1}{2} (y-2)^2-2\]
Vertex (-2,2)

for focus, the coefficient of the unsquared part is -2, 4p=-2, p=-1/2
focus=-1/2 + (-2)=-5/2

Focus (-5/2,2)

Answer 5

Do you mind if I ask another one on here? @.Sam.

Answer 6

yes?

Answer 7

y^2+6y+8x-7=0
How do you graph this parabola?

Answer 8

arrange it like this,
\[x=-\frac{y^2}{8}-\frac{3 y}{4}+\frac{7}{8}\]
complete the square
\[x= -\frac{1}{8} (y+3)^2+2\]
can you find the vertex and foci from here?

Answer 9

Well I completed the square instead of doing it that way. But how do you graph it?

Answer 10

you got the vertex (2,-3), and foci, (0,-3)