Question

find the maximum value of the function $f(x) = -0.6x^2 - 3.7x + 1$ to the nearest hundredth.

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a=-0.6 \), \( b = - 3.7 \), \( c = 1 \).

Step2: Calculate x - coordinate of vertex

Substitute \( a=-0.6 \) and \( b=-3.7 \) into the formula: \( x=-\frac{-3.7}{2\times(-0.6)}=-\frac{3.7}{1.2}\approx - 3.0833 \)

Step3: Calculate maximum value (y - coordinate)

Substitute \( x = - 3.0833 \) into \( f(x)=-0.6x^{2}-3.7x + 1 \)
\( f(-3.0833)=-0.6\times(-3.0833)^{2}-3.7\times(-3.0833)+1 \)
First, calculate \( (-3.0833)^{2}\approx9.5069 \)
Then, \( -0.6\times9.5069\approx - 5.7041 \)
\( -3.7\times(-3.0833)\approx11.4082 \)
Now, \( f(-3.0833)=-5.7041 + 11.4082+1=6.7041\approx6.70 \) (to the nearest hundredth)

Answer:

\( 6.70 \)