Let's break down and solve each of the five functions given in the worksheet. I'll also explain how to find the domain and range of each.

1. $f(x) = -4x + 12$

This is a linear function, as it is in the form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

• Domain: A linear function has no restrictions, so the domain is all real numbers: $(-\infty, \infty)$.
• Range: Since the function is linear and can take any real value as $x$ changes, the range is also $(-\infty, \infty)$.

To graph this, the slope $m = -4$ means that for each unit increase in $x$, $y$ decreases by 4. The y-intercept is 12, meaning the graph crosses the y-axis at $(0, 12)$.

2. $g(x) = -7$

This is a constant function.

• Domain: Since it is defined for all $x$, the domain is $(-\infty, \infty)$.
• Range: The range is just the constant value $g(x) = -7$, so the range is $\{-7\}$.

The graph is a horizontal line at $y = -7$.

3. $h(x) = -4x^2 - 16x + 1$

This is a quadratic function. To find the domain and range, we will need to analyze its vertex form. We can complete the square or use the vertex formula to find the vertex.

• Domain: A quadratic function is defined for all real values of $x$, so the domain is $(-\infty, \infty)$.

• Range: The parabola opens downward (since the coefficient of $x^2$ is negative). To find the range, we need the vertex. The vertex formula is:

  $x = \frac{-b}{2a} = \frac{-(-16)}{2(-4)} = \frac{16}{-8} = -2$

  Substituting $x = -2$ back into the function to find the corresponding $y$-value:

  $h(-2) = -4(-2)^2 - 16(-2) + 1 = -4(4) + 32 + 1 = -16 + 32 + 1 = 17$

  So the vertex is $(-2, 17)$, and since the parabola opens downward, the range is $(-\infty, 17]$.

4. $r(x) = \sqrt{-2x + 6}$

This is a square root function. The expression inside the square root must be non-negative for the function to be defined.

• Domain: We solve the inequality $-2x + 6 \geq 0$:

  $-2x \geq -6 \quad \Rightarrow \quad x \leq 3$

  So the domain is $(-\infty, 3]$.

• Range: The square root function always produces non-negative outputs. Therefore, the range is $[0, \infty)$.

5. $f(x) = \sqrt{3x + 12}$

This is also a square root function. Again, the expression inside the square root must be non-negative.

• Domain: Solve $3x + 12 \geq 0$:

  $3x \geq -12 \quad \Rightarrow \quad x \geq -4$

  So the domain is $[-4, \infty)$.

• Range: Since the square root function yields non-negative values, the range is $[0, \infty)$.

Let me know if you need any further clarifications, and here are five related questions you might want to explore further:

1. How do you find the vertex of a quadratic function algebraically?
2. What does the slope of a linear function represent geometrically?
3. Why do square root functions have restricted domains and ranges?
4. How would you graph a quadratic function by completing the square?
5. How can transformations (like shifts and stretches) affect the graph of a square root function?

Tip: Always check if the function has any domain restrictions (like square roots or divisions) before finding its range.