What is the equation of the parabola that has a vertex at (11,6) and passes through point (13,36)?

1 Answer
Apr 17, 2017

y=596(x+11)2+6
or
y=596x2+5548x+118196

Explanation:

The standard form of a parabola is y=a(xh)2+k, where a is a constant, vertex is (h,k) and the axis of symmetry is x=h.

Solve for a by substituting h=11,k=6 & x=13,y=36:

36=a(13+11)2+6

36=576a+6

30=576a

a=30576=596

Equation in standard form is y=596(x+11)2+6

General form is y=Ax2+Bx+C

Distribute right side of the equation:
y=596(x2+22x+121)+6

y=596x2+5548x+60596+6

y=596x2+5548x+118196