find the minimum value of the function $f(x)=x^{2}+3.9x + 0.4$ to the nearest hundredth.

Explanation:

Step1: Identify the coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 1$, $b=3.9$, $c = 0.4$.

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 $a = 1$ and $b = 3.9$ into the formula: $x=-\frac{3.9}{2\times1}=-1.95$.

Step3: Find the minimum value

Substitute $x=-1.95$ into the function $f(x)=x^{2}+3.9x + 0.4$. $f(-1.95)=(-1.95)^{2}+3.9\times(-1.95)+0.4$ $=3.8025-7.605 + 0.4$ $=-3.4025\approx - 3.40$

Answer:

$-3.40$