The problem on the screen asks to "Write the standard form of the quadratic function that has the indicated vertex and passes through the given point."

Information given:

1. Vertex of the parabola: $(-6, 2)$
2. Point on the parabola: $(-4, 3)$

Solution Steps:

1. Vertex Form of a Quadratic Function: The vertex form of a quadratic function is: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex of the parabola.

2. Substitute Vertex Values: Since the vertex is $(-6, 2)$, we substitute $h = -6$ and $k = 2$: $f(x) = a(x + 6)^2 + 2$

3. Use the Point $(-4, 3)$ to Solve for $a$: Substitute $x = -4$ and $f(x) = 3$: $3 = a(-4 + 6)^2 + 2$ Simplify inside the parentheses: $3 = a(2)^2 + 2$ $3 = 4a + 2$ Subtract 2 from both sides: $1 = 4a$ Divide by 4: $a = \frac{1}{4}$

4. Write the Final Equation: Substitute $a = \frac{1}{4}$ back into the equation: $f(x) = \frac{1}{4}(x + 6)^2 + 2$

This is the standard form of the quadratic function with the given vertex and passing through the given point.

Would you like further details on any specific part of this solution?

Further Questions:

1. What is the process for converting from vertex form to standard form of a quadratic function?
2. How do you find the axis of symmetry from the vertex form of a quadratic function?
3. How would the equation change if the given point was different?
4. What are other methods for finding the value of $a$ in a quadratic function?
5. How can you verify the solution by graphing?

Tip:

Always double-check your calculations, especially when substituting points to solve for parameters.