The problem asks to determine the values of $a$ and $k$ for the function $y = a(x + 1)^2 + k$, given that two points $(1, 7)$ and $(2, -23)$ lie on the graph.

We can use these points by substituting them into the equation to create a system of equations.

Step 1: Plug in $(1, 7)$ into the equation.

For $x = 1$ and $y = 7$:

$7 = a(1 + 1)^2 + k$ $7 = a(2)^2 + k$ $7 = 4a + k \quad \text{(Equation 1)}$

Step 2: Plug in $(2, -23)$ into the equation.

For $x = 2$ and $y = -23$:

$-23 = a(2 + 1)^2 + k$ $-23 = a(3)^2 + k$ $-23 = 9a + k \quad \text{(Equation 2)}$

Step 3: Solve the system of equations.

We now have the system:

$7 = 4a + k \quad \text{(Equation 1)}$ $-23 = 9a + k \quad \text{(Equation 2)}$

Subtract Equation 1 from Equation 2 to eliminate $k$:

$(-23 - 7) = (9a - 4a)$ $-30 = 5a$ $a = -6$

Step 4: Substitute $a = -6$ back into one of the original equations.

Substitute into Equation 1:

$7 = 4(-6) + k$ $7 = -24 + k$ $k = 7 + 24 = 31$

Final Answer:

The values of $a$ and $k$ are:

$a = -6, \quad k = 31$

Thus, the equation of the function is:

$y = -6(x + 1)^2 + 31$

Would you like more details or have any other questions?

Here are 5 related questions for further practice:

1. How would the equation change if the vertex of the parabola shifted to a different point?
2. What does the value of $a$ represent in terms of the graph’s shape and direction?
3. If another point was given, how could you verify the solution?
4. How do changes in $k$ affect the position of the parabola?
5. What would the equation be if the parabola opened upwards?

Tip: When solving systems of equations, substitution or elimination methods are key strategies.