The hyperbola is given by the equation
[tex]\frac{x^2}{16}-\frac{y^2}{4}=1[/tex]Notice that this is a hyperbola centered in the origin. Furthermore, as the x^2 is positive and the y^2 is negative, we can know that the graph opens to the right and the left; therefore, we expect the lines that cross the vertices of the hyperbola to be in a line of the form:
[tex]x=k;k\text{ a constant}[/tex]To sum up, given the characteristics of the equation, the hyperbola looks like this:
We need to find the value of the blue lines. For this, set y=0 and solve the equation as follows:
[tex]\begin{gathered} y=0 \\ \Rightarrow\frac{x^2}{16}=1 \\ \Rightarrow x^2=16 \\ \Rightarrow x=\pm4 \end{gathered}[/tex]Therefore, the vertices are (4,0) and (-4,0). Option C is the answer