Question

1. find the maximum value of the function $f(x) = -1.1x^2 - 2.6x + 6$ to the nearest hundredth.

Explanation:

Step1: Identify the formula for vertex x-coordinate

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = -1.1 \), \( b = -2.6 \).
\( x = -\frac{-2.6}{2\times(-1.1)} = -\frac{2.6}{2.2} \approx -1.1818 \)

Step2: Substitute x into the function to find maximum

Substitute \( x \approx -1.1818 \) into \( f(x) = -1.1x^2 - 2.6x + 6 \).
\( f(-1.1818) = -1.1\times(-1.1818)^2 - 2.6\times(-1.1818) + 6 \)
First, calculate \( (-1.1818)^2 \approx 1.3966 \)
Then, \( -1.1\times1.3966 \approx -1.5363 \)
\( -2.6\times(-1.1818) \approx 3.0727 \)
Now, sum them up: \( -1.5363 + 3.0727 + 6 \approx 7.5364 \approx 7.54 \)

Answer:

7.54