find the minimum value of the function $f(x)=2x^{2}-4.5x - 5$ to the nearest hundredth.

Explanation:

Step1: Find the x - coordinate of the vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is given by $x=-\frac{b}{2a}$. Here, $a = 2$, $b=-4.5$, so $x=-\frac{-4.5}{2\times2}=\frac{4.5}{4}=1.125$.

Step2: Find the y - coordinate of the vertex

Substitute $x = 1.125$ into the function $f(x)=2x^{2}-4.5x - 5$. $f(1.125)=2\times(1.125)^{2}-4.5\times1.125 - 5$. First, $(1.125)^{2}=1.265625$, then $2\times(1.125)^{2}=2\times1.265625 = 2.53125$. $4.5\times1.125 = 5.0625$. $f(1.125)=2.53125-5.0625 - 5=-7.53125\approx - 7.53$.

Answer:

$-7.53$