All parabolas of the form
y=ax2 + bx + c
have their axis of symmetry passing through the vertex.
Thus a and b can be eliminated on that basis.
Note that when the x2 term is positive, the parabola has a minimum at the vertex.
By inspection, when x=2, the value of y is at its minimum because the 5(x-2)2 term is zero. Any other value of x will render the term > 0. Thus the vertex is at x=2, where y=3.
Thus the answer is (C).
Check:
c. y=5(2-2)2+3=3
d. y=5(-2-2)2+3=83 (not equal to 3)