The vertex is (-2,5) and the y intercept is(0,6)

Question

Plz tell me the procedures on getting this equation solved

Answer 1

vertex = (-2,5), y intercept (0,6)

upward opening parabola. vertex=minimum point

y=a(x-h)^2 +k (h,k)=(-2,5)

y=a(x+2)^2+5

6 =a(2^2)+5

4a=6-5=1

a=1/4

y=f(x)= (1/4)(x-2)^2 + 5

that's probably the equation you want.

but there's an infinite number of parabolas with that vertex and y intercept, including a more standard rightward opening parabola:

x = a(y-k)^2 + h

x =a(y-5)^2 -2

0 = a(6-5)^2 -2

a = 2

x= 2(y-5)^2 -2

or

(x+2)/2 = (y-5)^2

y-5 = +/-sqr((x+2)/2)

y = 5+/-sqr[x/2 +1]

but this and other non-vertical parabolas are not functions, as they fail the vertical line test

parabolas at an angle will have an xy term, still a quadratic equation, a parabola, just not vertical or horizontal, and also not a function.