Graph the parabola and give its vertex, axis of symmetry. $x$-intercepts, and $y$-intercept $y=x^{2}-7 x+10$ The vertex is $\square$. (Type an ordered pair)

Step 1 ：The given equation is in the form \(y=ax^2+bx+c\), where \(a=1\), \(b=-7\), and \(c=10\).

Step 2 ：We need to convert this equation to the vertex form, which is \(y=a(x-h)^2+k\).

Step 3 ：The vertex \((h,k)\) can be found using the formulas \(h=-\frac{b}{2a}\) and \(k=c-\frac{b^2}{4a}\).

Step 4 ：Substituting the values of \(a\), \(b\), and \(c\) into these formulas, we get \(h=3.5\) and \(k=-2.25\).

Step 5 ：Final Answer: The vertex of the parabola is \(\boxed{(3.5, -2.25)}\).