Answer:
the third option (a=1, h=0, k=6)
Explanation:
if I understand correctly what your teacher wants from you, then you need find a (the factor of x² in the equation) and the vertex (turnaround point) of the parabola represented by such a quadratic equation.
the vertex point coordinates are called (h, k).
the general form of such an equation equation is
y = ax² + bx + c
so, we have a right away : a=1
now we can make this quickly by using common sense, or a bit more complex by going through mathematical formulas.
the fast, practical way is to know that y=x² is the very basic parabola with its vertex at (0, 0).
y = x² + 6 is simply the same parabola just lifted up (y direction) by 6 units, that makes the vertex (0, 6).
in pure theory, though, we need to find the transformation from the general y = ax² + bx + c form to
y = a(x - h)² + k
we see right away that k = f(h) = h² + 6
y = a(x² - 2xh + h²) + k
a=1
y = x² - 2xh + h² + k = x² - 2xh + h² + h² + 6 =
= x² - 2xh + 2h² + 6
comparing with y = x² + 6
we know that
-2×h = 0
as we have no term with just x.
=> h = 0
2h² + 6 = k
2×0² + 6 = 6 = k