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

Zeros: $\dfrac{3}{4}, -\dfrac{1}{4}$ $x$-intercepts: $\dfrac{3}{4}, -\dfrac{1}{4}$
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 $g(x)=0$: $$x^2-\dfrac{1}{2}x-\dfrac{3}{16}=0$$ Rearranging the Equation: $$x^2-\dfrac{1}{2}x=\dfrac{3}{16}$$ The coefficient of $x^2$ is $1$ and that of $x$ is $-\dfrac{1}{2}$, complete the square by adding $\left(\dfrac{1 \times -0.5}{2}\right)^2 = \dfrac{1}{16}$ $$\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{3}{16}+\dfrac{1}{16}$$ $$\left(x-\dfrac{1}{4}\right)^2 = \dfrac{1}{4}$$ $$\left(x-\dfrac{1}{4}\right)=\pm \dfrac{1}{2}$$ $\therefore x-\dfrac{1}{4} = \dfrac{1}{2} \hspace{20pt}\text{ or } \hspace{20pt} x-\dfrac{1}{4} = -\dfrac{1}{2}$ Solve each equation to obtain the zeros of the functoin: $x-\dfrac{1}{4} = \dfrac{1}{2} \to x=\dfrac{3}{4}$ $x-\dfrac{1}{4} = -\dfrac{1}{2} \to x= -\dfrac{1}{4}$ The $x$-intercepts of the function are $\dfrac{3}{4}$ and $-\dfrac{1}{4}$.