Answer:
Explanation:
Simply by comparing the given
f(x)=(x-2)^2-3 to
f(x) = (x-h)^2 + k, we see that h = 2 and k = -3, which tells us that the vertex of the graph is (2, -3). This parabola opens up because the coefficient of (x-2)^2 is +1.
Evaluating f(x)=(x-2)^2-3 at x = 4 (an arbitrary value), we see that
f(4) = (4-2)^2 - 3 = 4 - 3 = 1.
The point (4, 1) is also on the graph of this parabola.
Graph the vertex (2, -3) and the arbitrarily chosen point (4, 1). Remember that (2, -3) is the minimum of this function, so for x other than 2, the y-value is greater than -3.