Question

1. find the maximum value of the function f(x) = -1.1x² - 2.6x + 6 to the nearest hundredt

Explanation:

Step1: Identify the formula for vertex of a parabola

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

Step2: Substitute x into the function to find the maximum value

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 \( (-1.1818)^{2}\approx1.3966 \)
Then, \( -1.1\times1.3966\approx - 1.5363 \)
\( -2.6\times(-1.1818)\approx3.0727 \)
Now, \( f(-1.1818)=-1.5363 + 3.0727+6 \)
\( f(-1.1818)=-1.5363+9.0727 = 7.5364\approx7.54 \)

Answer:

The maximum value of the function \( f(x)=-1.1x^{2}-2.6x + 6 \) to the nearest hundredth is \( 7.54 \)