find the minimum value of the function f(x)=2x² - 6x + 8.3 to the nearest hundredth.

Explanation:

Step1: Identify the coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 2$, $b=-6$, $c = 8.3$.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function is given by $x=-\frac{b}{2a}$. Substitute $a = 2$ and $b=-6$ into the formula: $x=-\frac{-6}{2\times2}=\frac{6}{4}=1.5$.

Step3: Find the minimum value

Substitute $x = 1.5$ into the function $f(x)=2x^{2}-6x + 8.3$. $f(1.5)=2\times(1.5)^{2}-6\times1.5 + 8.3$. First, calculate $(1.5)^{2}=2.25$. Then $2\times(1.5)^{2}=2\times2.25 = 4.5$. And $6\times1.5 = 9$. So $f(1.5)=4.5-9 + 8.3=3.8$.

Answer:

$3.80$