## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Zeros: $3\sqrt{2}+3,-3\sqrt{2}+3$ $x$-intercepts: $3\sqrt{2}+3,-3\sqrt{2}+3$
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 $3\sqrt2 +3$.