Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the parabola's axis of symmetry. Use the graph to determine the function's domain and range.
\[
f(x)=x^{2}+10 x+5
\]

Step 1 ：The vertex form of a quadratic function is given by \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. The axis of symmetry is the line \(x = h\). The domain of a quadratic function is all real numbers, and the range is all numbers greater than or equal to the y-coordinate of the vertex if the parabola opens upwards \(a > 0\), or all numbers less than or equal to the y-coordinate of the vertex if the parabola opens downwards \(a < 0\).

Step 2 ：To find the vertex, we can complete the square on the given quadratic function. The x-coordinate of the vertex is given by \(-b/2a\), and the y-coordinate is \(f(-b/2a)\). The intercepts are the x-values for which \(f(x) = 0\).

Step 3 ：Calculating the vertex, we find that \(x = -5\) and \(f = -20\), so the vertex is \((-5, -20)\).

Step 4 ：The intercepts are at \(x = -5 - 2\sqrt{5}\) and \(x = -5 + 2\sqrt{5}\).

Step 5 ：The axis of symmetry is the line \(x = -5\).

Step 6 ：The domain of the function is all real numbers, and the range is all numbers greater than or equal to -20.

Step 7 ：\(\boxed{\text{Final Answer: The vertex of the parabola is at } (-5, -20). \text{ The intercepts are at } x = -5 - 2\sqrt{5} \text{ and } x = -5 + 2\sqrt{5}. \text{ The axis of symmetry is the line } x = -5. \text{ The domain of the function is all real numbers, and the range is all numbers greater than or equal to -20.}}\)