Given the general form of a parabola x2+6x-28y+11=0, what is the y coordinate the vertex? k=_____. (round to 3 decimal places)

Answer 1

x^2 +6x + 11 -28y =0

solve for y

y= (1/28)(x^2 +6x +11)

complete the square

y = (1/28)(x^2 + 6x+9) + 11/28- 9/28

y = (1/28)x+3)^2 + 2/28

y = (1/28)(x+3)^2 + 1/14

y = a(x-h)^2 + k is vertex form of a quadrtic equation

(h,k) = vertex = (-3, 1/14)

or use calculus

y=x^2/28 + 6x/28 + 11/28

y' = x/14 + 3/14 = 0

x+3=0

x=-3 = x coordinate of the vertex

y = 3^2/28 -6(3)/28 + 11/28

= 9/28 -18/28 +11/28 = 2/28

y = 1/14 = y coordinate of the vertex

h=-3, k=1/14 = y coordinate of the vertex