Let's go through each part of the question based on the function $Q(t) = 12 - 11t$.

(a) Lies six units above the graph of $Q(t)$:

To create a function that lies six units above $Q(t)$, we simply add 6 to $Q(t)$: $y(t) = Q(t) + 6 = (12 - 11t) + 6 = 18 - 11t$

(b) Lies two units below the graph of $Q(t)$:

To create a function that lies two units below $Q(t)$, we subtract 2 from $Q(t)$: $y(t) = Q(t) - 2 = (12 - 11t) - 2 = 10 - 11t$

(c) Has the same vertical intercept:

The vertical intercept is the value of the function when $t = 0$. For $Q(t) = 12 - 11t$, the intercept is 12. Any function with the same vertical intercept will have a constant of 12. So, the answer is any function of the form: y(t) = 12 + mt \quad \text{(where m is any slope)}

(d) Has the same slope:

The slope of $Q(t)$ is the coefficient of $t$, which is -11. A function with the same slope must have this same $t$-coefficient. The general form would be: $y(t) = mt + b \quad \text{where} \, m = -11$ So, the answer is: y(t) = -11t + b \quad \text{(any constant b)}

(e) Has the same steepness, but the slope is positive:

The "steepness" refers to the absolute value of the slope. For $Q(t)$, the steepness is $| -11 | = 11$, but we want the slope to be positive. The function will have the same steepness but a positive slope: y(t) = 11t + b \quad \text{(any constant b)}

Would you like more details on any of these? Here are five related questions to further explore the concepts:

1. How do we determine the slope and intercept of a linear function from its equation?
2. What transformations (like translations) can be applied to linear functions?
3. How does changing the slope affect the graph of a function?
4. Can two different functions have the same slope but different intercepts?
5. How do vertical shifts in a graph relate to addition or subtraction of constants?

Tip: When shifting a graph vertically, you simply add or subtract a constant to the entire function.