Let's solve problems 2 and 14 from the image:

Problem 2:

Given: $f(y) = y^2 + y - 6$ Find:

1. $f(2)$
2. $f(-1)$
3. $f(0)$

Solution:

1. For $f(2)$: $f(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$
2. For $f(-1)$: $f(-1) = (-1)^2 + (-1) - 6 = 1 - 1 - 6 = -6$
3. For $f(0)$: $f(0) = 0^2 + 0 - 6 = -6$

Problem 14:

Given interval: $(-1.5, 4.5)$

Task: Write the inequality representation of the interval.

Solution: The interval $(-1.5, 4.5)$ can be represented as an inequality: $-1.5 < x < 4.5$

Would you like further explanations or have any questions?

Follow-Up Questions:

1. How do you solve for $f(x)$ when given different values of $x$?
2. How do you convert intervals into inequalities?
3. What are open and closed intervals in interval notation?
4. How would the solution change if $x$ were included in the endpoints for Problem 14?
5. Can you explain why $f(0)$ and $f(-1)$ have the same value in Problem 2?

Tip: Always substitute the given values correctly in functions, and remember to follow the order of operations (PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).