question find the minimum value of the function f(x)=0.6x² - 1.5x + 8.1 to the nearest hundredth. answer…

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 0.6$, $b=-1.5$, $c = 8.1$.

Step2: Find x - coordinate of vertex

The x - coordinate of the vertex of a quadratic function is given by $x=-\frac{b}{2a}$. Substitute $a = 0.6$ and $b=-1.5$ into the formula: $x=-\frac{-1.5}{2\times0.6}=\frac{1.5}{1.2}=1.25$.

Step3: Find the minimum value

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

Answer:

$7.16$