Answer
Zeros: $3\sqrt{2}+3,-3\sqrt{2}+3$
$x$-intercepts: $3\sqrt{2}+3,-3\sqrt{2}+3$
Work Step by Step
To find the zeros of a function $g$, solve the equation $g(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $f(x)=0$:
$$x^2-6x-9=0$$
Rearranging the Equation:
$$x^2-6x=9$$
The coefficient of $x^2$ is 1 and that of x is -6, complete the square by adding $\left(\dfrac{1 \times -6}{2}\right)^2 = 9$ to both sides:
$$(x^2-6x+9)=9+9$$
$$(x-3)^2 = 18$$
$$(x-3)=\pm \sqrt{18}$$
Note that :
$\sqrt{18}= \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$
$\therefore x-3 = 3\sqrt{2} \hspace{20pt}\text{ or } \hspace{20pt} x-3 = -3\sqrt{2} $
Solve each equation to obtain the zeros of the function:
$x-3 = 3\sqrt{2} \to x=3\sqrt{2}+3$
$x-3 = -3\sqrt{2} \to x= -3\sqrt{2}+3$
Thus, the $x$-intercepts are $-3\sqrt2+3$ and $3\sqrt2 +3$.
