Question

x^2+4x-4y+16=0
-put in standard form
-find vertex
-find focus

Answer 1

then 4y = x^2 +4x + 16

complete the square for x

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

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

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

now in the standard form

vertex is at (-2, 3)

Answer 2

x^2 = 4ay

in this question

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

so a= 1

focus is at (-2, 4)

Answer 3

so whats the final answer in standard form?

Answer 4

which standard form

\[(x - h)^2 = 4a(y-k) \]

or

\[y =ax^2+bx + c\]

Answer 5

and the vertex and focus don't have a standard form