\[
h(x)=-x^{2}+4 x-3
\]
: Find two points on the graph of the parabola other than the vertex and $x$-intercepts.
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Step 1 ：The vertex of the parabola is given by the formula \(x = -\frac{b}{2a}\), where \(a\) and \(b\) are the coefficients of \(x^2\) and \(x\) respectively. In this case, \(a = -1\) and \(b = 4\), so the x-coordinate of the vertex is \(x = -\frac{4}{2(-1)} = 2\).

Step 2 ：The x-intercepts of the parabola are found by setting \(h(x) = 0\) and solving for \(x\). So we have \(-x^2 + 4x - 3 = 0\). Solving this quadratic equation, we get \(x = 1\) and \(x = 3\).

Step 3 ：So we cannot choose \(x = 1\), \(x = 2\), or \(x = 3\). Let's choose \(x = 0\) and \(x = 4\).

Step 4 ：For \(x = 0\), we substitute into the equation to get \(h(0) = -(0)^2 + 4(0) - 3 = -3\).

Step 5 ：For \(x = 4\), we substitute into the equation to get \(h(4) = -(4)^2 + 4(4) - 3 = -7\).

Step 6 ：\(\boxed{\text{So the two points on the graph of the parabola other than the vertex and x-intercepts are }(0, -3)\text{ and }(4, -7)}\)