what is the minimum value of the function $f(x)=2x^{2}-4x - 6$? enter the answer in the box.

Explanation:

Step1: Identify the coefficients

For the quadratic function $f(x)=2x^{2}-4x - 6$, where $a = 2$, $b=-4$, $c=-6$.

Step2: Find the x - coordinate of the vertex

The formula for the x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a = 2$ and $b=-4$ into the formula: $x=-\frac{-4}{2\times2}=\frac{4}{4}=1$.

Step3: Find the minimum value

Substitute $x = 1$ into the function $f(x)=2x^{2}-4x - 6$. Then $f(1)=2\times1^{2}-4\times1 - 6=2 - 4-6=-8$.

Answer:

-8