Answer:
[tex](x-2)^2=4(y+5)[/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 = (2, -5)
- Focus = (2, -4)
- Directrix: y = -6
- Axis of symmetry: x = 2
Therefore:
- h = 2
- k = -5
- k + p = -4 ⇒ -5 + p = -4 ⇒ p = 1
- k - p = -6 ⇒ -5 - p = -6 ⇒ p = 1
Substitute the found values of h, k and p into the formula:
[tex]\implies (x-2)^2=4(1)(y-(-5))[/tex]
[tex]\implies (x-2)^2=4(y+5)[/tex]
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