Answer
Zeros: $2+\sqrt{2}, 2-\sqrt{2}$
$x$-intercepts: $2+\sqrt{2}, 2-\sqrt{2}$
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 $f(x)=0$:
$$x^2-4x+2=0$$
Comparing $x^2-4x+2=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 1, b=-4 , c =2$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (-4)^2-4 \times 1 \times 2 = 8$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{4 \pm \sqrt{8}}{2\times 1}$$
$$x=\dfrac{4 \pm 2\sqrt{2}}{2}$$
$$x= 2 \pm \sqrt{2}$$
$\therefore x =2+\sqrt{2} \hspace{20pt} \text{or} \hspace{20pt} x=2-\sqrt{2}$
