The image shows a graph of a quadratic function $f(x) = ax^2 + bx + c$ and asks to determine the values of $a$, $b$, and $c$. From the graph, we observe a parabola that opens downwards and has the following key points:

• Vertex at $(2, 4)$
• The x-intercepts at $x = 0$ and $x = 4$

Step-by-Step Solution

1. Determine the equation format:
   The vertex form of a quadratic function is: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex. Here, the vertex is $(2, 4)$, so: $f(x) = a(x - 2)^2 + 4$

2. Use x-intercepts to find $a$:
   The x-intercepts are at $x = 0$ and $x = 4$, so: $f(0) = 0 \quad \text{and} \quad f(4) = 0$

   Substitute $x = 0$ into the vertex form: $0 = a(0 - 2)^2 + 4 \implies 0 = a(4) + 4 \implies -4 = 4a \implies a = -1$

3. Find the standard form coefficients:

   Now that we have $a = -1$, substitute $a$ back into the vertex form: $f(x) = -1(x - 2)^2 + 4$

   Expand to convert it to standard form: $f(x) = -1(x^2 - 4x + 4) + 4$ $f(x) = -x^2 + 4x - 4 + 4$ $f(x) = -x^2 + 4x$

   Therefore, the coefficients are: $a = -1, \; b = 4, \; c = 0$

Conclusion

• $a = -1$
• $b = 4$
• $c = 0$

Would you like more details, or have any questions?

Here are 5 related questions to consider:

1. How do you find the vertex of a parabola from its equation?
2. What is the axis of symmetry for the quadratic function given?
3. How does changing the value of $a$ affect the graph of a quadratic function?
4. What is the significance of the discriminant in quadratic equations?
5. Can you convert any quadratic function into vertex form?

Tip: Practice plotting different quadratic functions to understand how each coefficient affects the shape and position of the parabola.