To determine the values of $a$, $b$, and $c$ for the quadratic function $y = a(x - b)^2 + c$, we analyze the given information:

Step 1: Identify the roots

The quadratic function has roots at $x = -2$ and $x = 6$, as it passes through $(-2, 0)$ and $(6, 0)$. The midpoint of the roots gives the $x$-coordinate of the vertex: $b = \frac{-2 + 6}{2} = 2.$ So, the vertex is at $x = 2$, and the equation becomes: $y = a(x - 2)^2 + c.$

Step 2: Use the maximum $y$-value

The maximum value of $y$ is $48$, which occurs at the vertex $(2, c)$. Therefore: $c = 48.$ Now the equation becomes: $y = a(x - 2)^2 + 48.$

Step 3: Use the roots to solve for $a$

The function passes through the roots $(-2, 0)$ and $(6, 0)$. Substituting $(-2, 0)$ into the equation: $0 = a((-2) - 2)^2 + 48.$ Simplify: $0 = a(-4)^2 + 48.$ $0 = 16a + 48.$ $16a = -48 \quad \Rightarrow \quad a = -3.$

Final Answer:

The quadratic function is: $y = -3(x - 2)^2 + 48,$ and the values are: $a = -3, \quad b = 2, \quad c = 48.$

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Related Questions:

1. How do we verify that the vertex is a maximum and not a minimum?
2. What are the axis of symmetry and domain of this quadratic function?
3. How would the function change if the maximum value were different?
4. What is the effect of changing $a$ on the graph of a quadratic function?
5. How do we derive the vertex formula for a quadratic function?

Tip:

For any quadratic function, the vertex can be found using the formula $x = -\frac{b}{2a}$ or the midpoint of the roots, depending on the given form of the equation!