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