Question 932594
Find the equation of a parabola that has a Minimum turning point at (2,-5). 
y = a(x-2)^2 - 5  for a = 1 (which is one such parabola)
y = (x-2)^2 - 5
y = x^2 -4x + 4 - 5
y = x^2 - 4x - 1
{{{drawing(300,300,   -6, 6, -6, 6, grid(1), blue(line(2,6,2,-6)) ,

circle(2,-5, 0.3),

graph( 300, 300, -6, 6, -6, 6,0, x^2-4x-1) )}}}