Graph the quadratic function $f(x)=x^{2}-14 x+48$. Give the (a) vertex, (b) axis, (c) domain, and (d) range. Then determine (e) the largest open interval of the domain over which the function is increasing and (f) the largest open interval over which the function is decreasing.

Step 1 ：First, we need to find the vertex of the quadratic function. The vertex of a quadratic function given in the form \(f(x)=ax^2+bx+c\) is at the point \((h,k)\), where \(h=-\frac{b}{2a}\) and \(k=f(h)\).

Step 2 ：For the given function \(f(x)=x^{2}-14x+48\), we have \(a=1\) and \(b=-14\). So, \(h=-\frac{-14}{2*1}=7\).

Step 3 ：Substitute \(x=7\) into the function to get \(k=f(7)=7^{2}-14*7+48=7\). So, the vertex of the function is at \((7,7)\).

Step 4 ：Next, we find the axis of symmetry. The axis of symmetry of a quadratic function is the vertical line \(x=h\), where \(h\) is the x-coordinate of the vertex. So, the axis of symmetry is \(x=7\).

Step 5 ：The domain of a quadratic function is all real numbers, so the domain is \(x \in \boxed{(-\infty,\infty)}\).

Step 6 ：The range of a quadratic function given in the form \(f(x)=ax^2+bx+c\), where \(a>0\), is \([k,\infty)\), where \(k\) is the y-coordinate of the vertex. Since the coefficient of \(x^2\) in the given function is 1, which is greater than 0, the range is \(y \in \boxed{[7,\infty)}\).

Step 7 ：The function is increasing on the interval \((h,\infty)\), where \(h\) is the x-coordinate of the vertex. So, the function is increasing on the interval \(x \in \boxed{(7,\infty)}\).

Step 8 ：The function is decreasing on the interval \((-\infty,h)\), where \(h\) is the x-coordinate of the vertex. So, the function is decreasing on the interval \(x \in \boxed{(-\infty,7)}\).