The image shows a problem asking for the maximum value of the quadratic function $f(x) = -0.6x^2 - 3.7x + 1$ to the nearest hundredth.

Step 1: Determine the Vertex of the Quadratic Function

For the quadratic function $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex can be calculated using: $x = -\frac{b}{2a}$ Given:

• $a = -0.6$
• $b = -3.7$

Substituting these values into the formula: $x = -\frac{-3.7}{2(-0.6)} = \frac{3.7}{-1.2} \approx -3.0833$

Step 2: Find the Maximum Value of the Function

Substitute $x \approx -3.0833$ back into the function to find the corresponding y-coordinate, which is the maximum value since the parabola opens downwards (because $a$ is negative): $f(-3.0833) = -0.6(-3.0833)^2 - 3.7(-3.0833) + 1$ $f(-3.0833) \approx -0.6(9.5088) + 11.2082 + 1$ $f(-3.0833) \approx -5.7053 + 11.2082 + 1 \approx 6.5029$

Rounding to the nearest hundredth: $f(-3.0833) \approx 6.50$

Conclusion

The maximum value of the function $f(x) = -0.6x^2 - 3.7x + 1$ is approximately 6.50.

Would you like further explanation or have any questions?

Here are 5 related questions to further your understanding:

1. Why does the function $f(x) = -0.6x^2 - 3.7x + 1$ have a maximum value instead of a minimum?
2. How does the value of $a$ affect the steepness of the parabola?
3. What is the significance of the vertex in determining the extremum of a quadratic function?
4. Can you find the x-intercepts of the function $f(x) = -0.6x^2 - 3.7x + 1$?
5. How do the values of $b$ and $c$ influence the position of the vertex on the graph?

Tip: Always double-check your calculations when rounding to ensure accuracy, especially in exams or assessments where precision matters.