find the minimum value of the function $f(x)=2x^{2}+24x + 74.4$ to the nearest hundredth.

Explanation:

Step1: Identify the coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 2$, $b=24$, $c = 74.4$.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substitute $a = 2$ and $b = 24$ into the formula: $x=-\frac{24}{2\times2}=-\frac{24}{4}=- 6$.

Step3: Find the minimum value of the function

Substitute $x=-6$ into the function $f(x)=2x^{2}+24x + 74.4$. $f(-6)=2\times(-6)^{2}+24\times(-6)+74.4$. First, calculate $2\times(-6)^{2}=2\times36 = 72$. Then, $24\times(-6)=-144$. So, $f(-6)=72-144 + 74.4=-144+146.4 = 2.4$.

Answer:

$2.40$