Answer:
[tex](x+3)^2=-4(y-3)[/tex]
Step-by-step explanation:
Standard form of a parabola with a vertical axis of symmetry
[tex]\large\boxed{(x-h)^2=4p(y-k) \quad \textsf{where}\:p\neq 0}[/tex]
[tex]\textsf{Vertex}=(h, k)[/tex]
[tex]\textsf{Focus}=(h,k+p)[/tex]
[tex]\textsf{Directrix}:y=(k-p)[/tex]
[tex]\textsf{Axis of symmetry}:x=h[/tex]
If p > 0, the parabola opens upwards.
If p < 0, the parabola opens downwards.
From inspection of the given graph:
- Vertex = (-3, 3)
- Focus = (-3, 2)
- Directrix: y = 4
- Axis of symmetry: x = -3
Therefore:
- h = -3
- k = 3
- k + p = 2 ⇒ 3 + p = 2 ⇒ p = -1
- k - p = 4 ⇒ 3 - p = 4 ⇒ p = -1
Substitute the found values of h, k and p into the formula:
[tex]\implies (x-(-3))^2=4(-1)(y-3)[/tex]
[tex]\implies (x+3)^2=-4(y-3)[/tex]
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