To analyze the given problem, let's observe the quadratic function graph provided in the image.

Analyzing the Information:

• The graph has its vertex at the point $(1, 6)$. Since the parabola opens downward, the leading coefficient of the quadratic function must be negative.
• The x-intercepts appear to be at $-2$ and $3$.

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. Substituting $(h, k) = (1, 6)$, the equation becomes: $f(x) = a(x - 1)^2 + 6$

Finding the Correct Equation:

To find the correct value of $a$, we can use the x-intercepts ($-2$ and $3$) to form the equation. Expanding this information into the standard form of a quadratic will lead us to: $f(x) = -x^2 + 2x + 6$ Based on the available options, let's compare this with each given option.

Given Options:

A. $f(x) = 6 - x - x^2$
B. $f(x) = -6 + x + x^2$
C. $f(x) = -6 + x + x^2$
D. $f(x) = 6 + x - x^2$

The correct choice is $f(x) = 6 + x - x^2$, which matches the characteristics of the given graph.

Final Answer:

The correct answer is option D: $f(x) = 6 + x - x^2$.

Would you like more details or have any questions on how this solution was reached?

Related Questions:

1. How do you convert a quadratic function from vertex form to standard form?
2. What are the steps to find the vertex of a quadratic function using its equation?
3. How can you determine the direction of a parabola from its equation?
4. How do you calculate the x-intercepts of a quadratic equation?
5. What role does the leading coefficient play in the shape of a parabola?

Tip:

Always check the sign of the leading coefficient in a quadratic equation to determine the direction in which the parabola opens.