Question

find the minimum value of the function $f(x) = 0.6x^2 - 1.5x + 8.1$ to the nearest hundredth.

Explanation:

Step1: Identify the coefficients of the quadratic function

For a quadratic function \( f(x) = ax^2 + bx + c \), here \( a = 0.6 \), \( b = -1.5 \), \( c = 8.1 \). The x - coordinate of the vertex of a parabola (which gives the minimum value for \( a>0 \)) is given by \( x = -\frac{b}{2a} \).

Step2: Calculate the x - coordinate of the vertex

Substitute \( a = 0.6 \) and \( b = -1.5 \) into the formula \( x = -\frac{b}{2a} \).
\( x=-\frac{-1.5}{2\times0.6}=\frac{1.5}{1.2} = 1.25 \)

Step3: Calculate the minimum value of the function

Substitute \( x = 1.25 \) into the function \( f(x)=0.6x^{2}-1.5x + 8.1 \).
\( f(1.25)=0.6\times(1.25)^{2}-1.5\times(1.25)+8.1 \)
First, calculate \( (1.25)^{2}=1.5625 \)
Then, \( 0.6\times1.5625 = 0.9375 \)
\( 1.5\times1.25 = 1.875 \)
So, \( f(1.25)=0.9375-1.875 + 8.1 \)
\( f(1.25)=0.9375+6.225=7.1625\approx7.16 \)

Answer:

\( 7.16 \)