The vertex is the highest or lowest point on the parabola. In this case, the vertex is (2, 1).
We can choose the x-intercept (3, 0).
The vertex form of a parabola is $y = a(x-h)^2 + k$, where (h, k) is the vertex. Plugging in the vertex (2, 1), we get $y = a(x-2)^2 + 1$.
Substitute the other point (3, 0) into the equation: $0 = a(3-2)^2 + 1$. Simplifying this, we get $0 = a(1)^2 + 1$, so $0 = a + 1$. Solving for $a$, we get $a = -1$.
Substitute $a = -1$ back into the vertex form: $y = -(x-2)^2 + 1$.
Expand the equation:
$y = -(x^2 - 4x + 4) + 1$
$y = -x^2 + 4x - 4 + 1$
$y = -x^2 + 4x - 3$