Answer
Zeros: $\dfrac{3}{4}, -\dfrac{1}{4}$
$x$-intercepts: $\dfrac{3}{4}, -\dfrac{1}{4}$
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 $g(x)=0$:
$$x^2-\dfrac{1}{2}x-\dfrac{3}{16}=0$$
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 $-\dfrac{1}{4}$.
