We are given the quadratic equation $y = x^2 + 4x - 1$ and are asked to complete the table for values of $y$ when $x = -3$, $-2$, $-1$, $0$, and $1$. Specifically, we need to find the values of $A$, $B$, and $C$.

Step-by-Step Calculations:

1. For $x = -3$: $y = (-3)^2 + 4(-3) - 1 = 9 - 12 - 1 = -4$ So, $A = -4$.

2. For $x = -2$: $y = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5$ This value is already provided in the table as $y = -5$.

3. For $x = -1$: $y = (-1)^2 + 4(-1) - 1 = 1 - 4 - 1 = -4$ So, $B = -4$.

4. For $x = 0$: $y = (0)^2 + 4(0) - 1 = -1$ This value is also provided in the table as $y = -1$.

5. For $x = 1$: $y = (1)^2 + 4(1) - 1 = 1 + 4 - 1 = 4$ So, $C = 4$.

Final Values:

• $A = -4$
• $B = -4$
• $C = 4$

Would you like more details on any part of this solution?

Here are some related questions to explore:

1. How do we derive the general formula for a quadratic equation?
2. What are the properties of the graph of a quadratic equation like $y = x^2 + 4x - 1$?
3. How can you find the vertex of a quadratic equation?
4. What is the axis of symmetry for this quadratic equation?
5. How do different values of $x$ affect the behavior of quadratic functions?

Tip: Always double-check your work by substituting values back into the original equation!