find the minimum value of the function $f(x)=1.5x^{2}+15.7x + 48$ to the nearest hundredth.

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 1.5$, $b=15.7$, $c = 48$.

Step2: Find x - coordinate of vertex

The x - coordinate of the vertex of a quadratic function is given by $x=-\frac{b}{2a}$. So $x=-\frac{15.7}{2\times1.5}=-\frac{15.7}{3}\approx - 5.233$.

Step3: Find the minimum value

Substitute $x =-\frac{15.7}{3}$ into the function $f(x)$. $f(-\frac{15.7}{3})=1.5\times(-\frac{15.7}{3})^{2}+15.7\times(-\frac{15.7}{3}) + 48$. First, $(-\frac{15.7}{3})^{2}=\frac{246.49}{9}$, then $1.5\times\frac{246.49}{9}=\frac{1.5\times246.49}{9}=\frac{369.735}{9}=41.08167$. $15.7\times(-\frac{15.7}{3})=-\frac{246.49}{3}\approx - 82.1633$. $f(-\frac{15.7}{3})=41.08167-82.1633 + 48=-82.1633+89.08167 = 6.91837\approx6.92$.

Answer:

$6.92$