Write the quadratic function in vertex form given the axis of symmetry and two non-vertex points A, O, S, colon, x, equals, minus, 2, comma, 1, s, t, P, o, i, n, t, colon, left parenthesis, minus, 3, comma, 2, right parenthesis, comma, 2, n, d, P, o, i, n, t, colon, left parenthesis, 0, comma, minus, 16, right parenthesisAOS:x=−2,1stPoint:(−3,2),2ndPoint:(0,−16)

We are tasked with finding the quadratic function in vertex form, given the axis of symmetry \( x = -2 \) and two points, \( (-3, 2) \) and \( (0, -16) \).

### Step-by-Step Solution:

1. **Vertex Form of a Quadratic Function**:
   The vertex form of a quadratic function is:
   \[
   y = a(x - h)^2 + k
   \]
   where \( (h, k) \) is the vertex of the parabola and \( a \) is a constant that determines the "width" and direction (upward or downward) of the parabola.

2. **Axis of Symmetry and Vertex**:
   The axis of symmetry is given as \( x = -2 \). In a quadratic function, the \( x \)-coordinate of the vertex is equal to the axis of symmetry, so the vertex is \( (-2, k) \), where \( k \) is an unknown constant that we will find.

   The equation becomes:
   \[
   y = a(x + 2)^2 + k
   \]

3. **Using the First Point \( (-3, 2) \)**:
   We can plug the point \( (-3, 2) \) into the equation to form a system of equations:
   \[
   2 = a(-3 + 2)^2 + k
   \]
   Simplifying:
   \[
   2 = a(1)^2 + k \implies 2 = a + k
   \]
   This gives us the equation:
   \[
   a + k = 2 \quad \text{(Equation 1)}
   \]

4. **Using the Second Point \( (0, -16) \)**:
   Similarly, substitute the second point \( (0, -16) \) into the equation:
   \[
   -16 = a(0 + 2)^2 + k
   \]
   Simplifying:
   \[
   -16 = a(2)^2 + k \implies -16 = 4a + k
   \]
   This gives us the equation:
   \[
   4a + k = -16 \quad \text{(Equation 2)}
   \]

5. **Solving the System of Equations**:
   We now have two equations:
   \[
   a + k = 2 \quad \text{(Equation 1)}
   \]
   \[
   4a + k = -16 \quad \text{(Equation 2)}
   \]
   Subtract Equation 1 from Equation 2:
   \[
   (4a + k) - (a + k) = -16 - 2
   \]
   Simplifying:
   \[
   3a = -18 \implies a = -6
   \]

6. **Find \( k \)**:
   Substitute \( a = -6 \) into Equation 1:
   \[
   -6 + k = 2 \implies k = 8
   \]

7. **Final Equation**:
   Now that we have \( a = -6 \) and \( k = 8 \), we can write the quadratic equation in vertex form:
   \[
   y = -6(x + 2)^2 + 8
   \]

### Conclusion:
The quadratic function in vertex form is:
\[
y = -6(x + 2)^2 + 8
\]

Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions to expand on this:

1. How do you convert a quadratic function from vertex form to standard form?
2. What is the significance of the constant \( a \) in the vertex form of a quadratic equation?
3. How do you find the vertex of a quadratic function if given the standard form?
4. Can you find the x-intercepts of this quadratic function from vertex form?
5. How do transformations of the parent function \( y = x^2 \) relate to vertex form?

**Tip**: Always check that the given points satisfy the final equation to ensure the solution is correct.

Write the quadratic function in vertex form given the axis of symmetry and two non-vertex points AOS: x = −2, 1st Point: (−3, 2), 2nd Point: (0, −16).

Learn how to find the quadratic function in vertex form given the axis of symmetry at x = -2 and two points (-3, 2) and (0, -16). This problem involves key algebraic concepts such as solving for a and k in the vertex form and solving a system of linear equations. Suitable for students in Grades 9-11.

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