I see you've uploaded a set of math problems, including topics like function domains, operations on functions, and graphing. Let's break these down one by one and address them.

1. Consider the function $f(x) = \sqrt{x^2 - 10} + 1$:

(a) Determine the domain and range of the function.
The expression under the square root, $x^2 - 10$, must be non-negative: $x^2 - 10 \geq 0 \implies x^2 \geq 10 \implies |x| \geq \sqrt{10}.$ Thus, the domain is $(-\infty, -\sqrt{10}] \cup [\sqrt{10}, \infty)$.

(b) Explain any restrictions on the domain and range.
The square root function restricts the domain as above. Also, since $f(x) \geq 1$, the range is $[1, \infty)$.

(c) Draw the graph of the function.
The graph is a modified version of $\sqrt{x^2 - 10}$, shifted upwards by 1 unit. You can sketch this by plotting points from the domain.

2. Given two functions $f(x) = x + 2$ and $g(x) = \frac{1}{x - 3}$, perform the following operations:

(a) Find $(f + g)(x)$ and simplify:
$(f + g)(x) = f(x) + g(x) = (x + 2) + \frac{1}{x - 3}.$ This is already simplified.

(b) Find $(f \circ g)(x)$ and simplify:
$(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x - 3}\right) = \frac{1}{x - 3} + 2.$

(c) Find $(g \circ f)(x)$ and simplify:
$(g \circ f)(x) = g(f(x)) = g(x + 2) = \frac{1}{(x + 2) - 3} = \frac{1}{x - 1}.$

(d) Determine the domain and range of the resulting functions from parts (a), (b), and (c):
For part (a), the domain is $(-\infty, 3) \cup (3, \infty)$ due to the restriction from $g(x)$.
For part (b), the domain is $(-\infty, 3) \cup (3, \infty)$.
For part (c), the domain is all real numbers except $x = 1$.

3. Consider the line passing through points A(3, -1) and B(-2, 4):

(a) Calculate the slope of the line passing through points A and B:
Using the slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{-2 - 3} = \frac{5}{-5} = -1.$

(b) Write the equation of the line in point-slope form:
The point-slope form is $y - y_1 = m(x - x_1)$. Using point $A(3, -1)$ and slope $m = -1$: $y + 1 = -1(x - 3).$

(c) Convert the equation from point-slope form to slope-intercept form:
Simplifying the equation from part (b): $y + 1 = -x + 3 \implies y = -x + 2.$

(d) Verify the intercepts by graphing the line:
The y-intercept is 2 (from the slope-intercept form $y = -x + 2$), and the x-intercept is found by setting $y = 0$: $0 = -x + 2 \implies x = 2.$

4. Draw the graph of ( f(x) = \begin{cases}

|x + 1|, & x < -5 \ (x - 1)^2 + 1, & -5 \leq x \leq 5 \ |x + 1|, & x \geq 5 \end{cases} ): This is a piecewise function that combines absolute value and quadratic components. You can plot this by evaluating each section:

• For $x < -5$, graph $|x + 1|$.
• For $-5 \leq x \leq 5$, graph the quadratic $(x - 1)^2 + 1$.
• For $x \geq 5$, graph $|x + 1|$.

Would you like any details or clarifications on these steps?

Here are 5 questions related to this material:

1. What happens to the range of $f(x)$ if you change the constant inside the square root function?
2. Can you identify the vertical asymptotes of $g(x) = \frac{1}{x - 3}$?
3. How would you interpret the slope geometrically in problem 3?
4. What other points can you use to verify the line equation in 3(d)?
5. How does the piecewise function in problem 4 behave as $x \to \infty$?

Tip: When working with piecewise functions, always check for continuity at the breakpoints between intervals.