To analyze the graph and solve for the quadratic function, we can start by understanding that this quadratic function, $f(x)$, follows the vertex form:

$f(x) = a(x - h)^2 + k$

where:

• $(h, k)$ is the vertex of the parabola.
• $a$ is the leading coefficient that determines the parabola's direction and width.

Part 1: Leading Coefficient

Since the vertex is at $(2, -16)$ and the parabola passes through $(0, 0)$, it opens downward, indicating that the leading coefficient, $a$, is negative.

Part 2: Values of $h$ and $k$

The vertex form of the function is $f(x) = a(x - h)^2 + k$, and the vertex is given as $(2, -16)$. Therefore:

• $h = 2$
• $k = -16$

Part 3: Finding the Leading Coefficient, $a$

To find $a$, we can substitute the point $(0, 0)$, which the parabola passes through, into the equation along with $h$ and $k$:

1. Substitute $(0, 0)$ into the function: $0 = a(0 - 2)^2 - 16$

2. Simplify and solve for $a$: $0 = a(4) - 16$ $16 = 4a$ $a = \frac{16}{4} = 4$

Since the parabola opens downward, $a = -4$.

Summary:

• a. The leading coefficient is negative.
• b. $h = 2$ and $k = -16$.
• c. The value of $a = -4$.

This gives us the equation for the function: $f(x) = -4(x - 2)^2 - 16$

Would you like more details or have any questions?

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1. What does changing the value of $a$ do to the shape of the parabola?
2. How can we verify that $(0, 0)$ lies on this parabola?
3. What happens if the vertex $(h, k)$ is at a different point?
4. How does the vertex form differ from the standard form of a quadratic function?
5. Why is the vertex form useful for identifying the maximum or minimum point?

Tip: When graphing quadratics, the vertex form is ideal for quickly identifying the vertex and direction of the parabola.