Consider the given function \( p(x)=x^{2}+10 x+16 \) (a) Write the function in vertex form. (b) Identify the vertex. (c) Determine the \( x \)-intercept(s). (d) Determine the \( y \)-intercept(s). (e) Sketch the function. (f) Determine the axis of symmetry. (g) Determine the minimum or maximum value of the function. (h) Write the domain and range in interval notation. Write your answers in exact form.

Let's tackle this step by step: (a) To write the function \( p(x) = x^{2} + 10x + 16 \) in vertex form, we can complete the square. Start by taking the \( x^{2} + 10x \) part: 1. Take half of the coefficient of \( x \) (which is 10), square it (which gives 25), and rewrite: \[ p(x) = (x^2 + 10x + 25) - 25 + 16 = (x + 5)^2 - 9. \] Thus, the vertex form is: \[ p(x) = (x + 5)^2 - 9. \] (b) The vertex can be read directly from the vertex form \( p(x) = (x + 5)^2 - 9 \). The vertex is at the point (-5, -9). (c) To find the \( x \)-intercepts, set \( p(x) = 0 \): \[ (x + 5)^2 - 9 = 0. \] This factors into \((x + 5 - 3)(x + 5 + 3) = 0\) or \((x + 2)(x + 8) = 0\), giving us: \[ x = -2, \, -8. \] Thus, the \( x \)-intercepts are \((-2, 0)\) and \((-8, 0)\). (d) To find the \( y \)-intercept, set \( x = 0 \): \[ p(0) = 0^{2} + 10(0) + 16 = 16. \] Thus, the \( y \)-intercept is \((0, 16)\). (e) To sketch the function, plot the vertex at (-5, -9), the \( y \)-intercept at (0, 16), and the \( x \)-intercepts at (-2, 0) and (-8, 0). The graph is a parabola opening upwards. (f) The axis of symmetry is given by the line \( x = -5 \), which is the \( x \)-coordinate of the vertex. (g) Since the parabola opens upwards, the minimum value of the function is the \( y \)-coordinate of the vertex, which is -9. (h) The domain of a quadratic function is all real numbers, so in interval notation, the domain is: \[ (-\infty, \infty). \] The range, since the vertex is the minimum point and the parabola opens upwards, is: \[ [-9, \infty). \]