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

Zeros: $\dfrac{1}{3},-1$ $x$-intercepts: $\dfrac{1}{3},-1$
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 $g(x)=0$: $$x^2+\dfrac{2}{3}x-\dfrac{1}{3}=0$$ Rearranging the Equation: $$x^2+\dfrac{2}{3}x=\dfrac{1}{3}$$ The coefficient of $x^2$ is 1 and that of x is \dfrac{2}{3}, complete the square by adding $\left(\dfrac{1 \times \dfrac{2}{3}}{2}\right)^2 = \dfrac{1}{9}$ $$\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}\right)=\dfrac{1}{3}+\dfrac{1}{9}$$ $$\left(x+\dfrac{1}{3}\right)^2 = \dfrac{4}{9}$$ $$\left(x+\dfrac{1}{3}\right)=\pm \dfrac{2}{3}$$ $\therefore x+\dfrac{1}{3} = \dfrac{2}{3} \hspace{20pt}\text{ or } \hspace{20pt} x+\dfrac{1}{3} = -\dfrac{2}{3}$ Solve each equation to obtain the zeros of the function: $x+\dfrac{1}{3} = \dfrac{2}{3} \to x=\dfrac{1}{3}$ $x+\dfrac{1}{3} = -\dfrac{2}{3} \to x= -1$ The $x$-intercepts of the function are $-1$ and $\dfrac{1}{3}$.