Question

Write an equation of the parabola that opens up whose vertex (−1, 2) is 3 units from the focus. (Vertex form or Parabola Form is acceptable) PLZZ HELP ASAP!!!!!!!!!!!

Answer 1

Answer:

y2  =  4ax (opens right, a > 0)

y2  =  -4ax (opens right, a > 0)

x2  =  4ay (opens up, a > 0)

x2  =  -4ay (opens down, a > 0)

Vertex at (h, k) :  

(y - k)2  =  4a(x - h) (opens right, a > 0)

(y - k)2  =  -4a(x - h) (opens right, a > 0)

(x - h)2  =  4a(y - k) (opens up, a > 0)

(x - h)2  =  -4a(y - k) (opens down, a > 0)

Equation of a Parabola in Vertex form

Vertex at Origin :  

y  =  ax2 (opens up, a > 0)

y  =  -ax2 (opens down, a > 0)

x  =  ay2 (opens right, a > 0)

x  =  -ay2 (opens left, a > 0)

Vertex at (h, k) :  

y  =  a(x - h)2 + k (opens up, a > 0)

y  =  -a(x - h)2 + k (opens down, a > 0)

x  =  a(y - k)2 + h (opens right, a > 0)

y  =  -a(y - k)2 + h (opens left, a > 0)

Step-by-step explanation: