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.

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**(a) Write the function in vertex form**

To convert 
$$
p(x)=x^2+10x+16
$$
into vertex form, we complete the square:

1. Factor the quadratic and linear terms:
   $$
   x^2+10x \quad (\text{we want to complete the square})
   $$

2. Find the number to complete the square:
   $$
   \left(\frac{10}{2}\right)^2 = 25.
   $$

3. Rewrite the function by adding and subtracting 25:
   $$
   p(x)=x^2+10x+25-25+16 = (x^2+10x+25) - 9.
   $$

4. Express the perfect square:
   $$
   x^2+10x+25=(x+5)^2.
   $$

Thus, the vertex form is:
$$
p(x)=(x+5)^2 - 9.
$$

**(b) Identify the vertex**

The vertex form $$(x+5)^2-9$$ shows that the vertex is at:
$$
(-5, \,-9).
$$

**(c) Determine the $$x$$-intercept(s)**

To find the $$x$$-intercepts, set $$p(x)=0$$:
$$
(x+5)^2-9=0.
$$
Solve for $$x$$:
$$
(x+5)^2=9
$$
Taking the square root of both sides:
$$
x+5=\pm 3.
$$

Thus,
- Case 1: $$x+5=3 \Rightarrow x= -2$$,
- Case 2: $$x+5=-3 \Rightarrow x= -8$$.

The $$x$$-intercepts are:
$$
(-2,0) \quad \text{and} \quad (-8,0).
$$

**(d) Determine the $$y$$-intercept(s)**

The $$y$$-intercept is found by evaluating $$p(0)$$:
$$
p(0)=0^2+10(0)+16=16.
$$
So, the $$y$$-intercept is:
$$
(0,16).
$$

**(e) Sketch the function**

The key features needed for the sketch are:
- **Vertex:** $$(-5, -9)$$
- **Axis of symmetry:** $$x=-5$$
- **$$x$$-intercepts:** $$(-8,0)$$ and $$(-2,0)$$
- **$$y$$-intercept:** $$(0,16)$$
- **Opening:** Since the coefficient of $$x^2$$ is positive ($$a=1$$), the parabola opens upward.

When sketching, plot the vertex, $$x$$-intercepts, $$y$$-intercept, and draw a symmetric parabola about the line $$x=-5$$.

**(f) Determine the axis of symmetry**

The axis of symmetry for a parabola in vertex form $$(x+5)^2-9$$ is the vertical line through the vertex. Thus, the axis of symmetry is:
$$
x = -5.
$$

**(g) Determine the minimum or maximum value of the function**

Since the parabola opens upward, the vertex represents the minimum point. The minimum value is the $$y$$-coordinate of the vertex:
$$
\text{Minimum value } = -9 \quad \text{at } x=-5.
$$

**(h) Write the domain and range in interval notation**

- **Domain:** All real numbers, so:
  $$
  (-\infty, \infty)
  $$
- **Range:** Since the minimum value is $$-9$$, the range is:
  $$
  [-9, \infty)
  $$
**(a) Vertex Form:** $$ p(x) = (x + 5)^2 - 9 $$ **(b) Vertex:** $$(-5, -9)$$ **(c) $$x$$-Intercepts:** $$(-8, 0)$$ and $$(-2, 0)$$ **(d) $$y$$-Intercept:** $$(0, 16)$$ **(e) Sketch:** A parabola opening upward with vertex at $$(-5, -9)$$, $$x$$-intercepts at $$(-8, 0)$$ and $$(-2, 0)$$, and $$y$$-intercept at $$(0, 16)$$. **(f) Axis of Symmetry:** $$x = -5$$ **(g) Minimum Value:** $$-9$$ at $$x = -5$$ **(h) Domain:** $$(-\infty, \infty)$$ **(h) Range:** $$[-9, \infty)$$

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