Answer
Zeros: $\sqrt{2}+2,\sqrt{2}-2$
$x$-intercepts: $\sqrt{2}+2,\sqrt{2}-2$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $F(x)=0$:
$$\dfrac{1}{2}x^2-\sqrt{2}x-1=0$$
Comparing $\dfrac{1}{2}x^2-\sqrt{2}x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = \dfrac{1}{2}, b=-\sqrt{2}, c =-1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (-\sqrt{2})^2-4 \times \dfrac{1}{2} \times -1 = 4$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{\sqrt{2}\pm \sqrt{4}}{2\times \dfrac{1}{2}}$$
$$x=\dfrac{\sqrt{2}\pm 2}{1}$$
Thus, the zeros, which are also the $x$-intercepts, are $\sqrt{2}+2$ and $\sqrt{2}-2$.
