We have the following factored form of the rational function, which is already simplified. \[ g(x)=\frac{6 x}{(x-1)(x+6)} \] Begin with the vertical asymptotes which are determined by the \( x \)-values that make the denominator of the simplified function zero. What are the equations of the vertical asymptotes? (Enter your answers as a comma-separated list of equations. If there are no asymptotes, enter DNE.)

Suppose that the functions \( f \) and \( g \) are defined as follows. \[ \begin{array}{l}f(x)=2 x+3 \\ g(x)=\sqrt{4 x+1} \\ \text { Find } \frac{f}{g} \text { and } f-g \text {. Then, give their domains using interval notation. } \\ \begin{array}{l}\left(\frac{f}{g}\right)(x)=\square \\ \text { Domain of } \frac{f}{g}: \square \\ \text { Domain of } f-g: \square\end{array} \\ \qquad\end{array} \] \( \begin{array}{l}\qquad-g)=\square\end{array} \) \[ \]

257. Funksiýalaryň grafigini guruń: \( \begin{array}{ll}\text { a) } y=x^{2}+6 x ; & \text { c) } y=0,5 x^{2}-4 \\ \text { b) } y=-\frac{1}{2} x^{2}+2 ; & \text { d) } y=\frac{1}{2} x^{2}-3 x\end{array} \)

257. Funksiýalaryň grafigini guruñ: \( \begin{array}{ll}\text { a) } y=x^{2}+6 x ; & \text { c) } y=0,5 x^{2}-4 \\ \text { b) } y=-\frac{1}{2} x^{2}+2 ; & \text { d) } y=\frac{1}{2} x^{2}-3 x\end{array} \)

Consider the following. \[ g(x)=\frac{x^{2}-2 x-35}{x^{2}-36} \] (a) State the domain of the function. (b) Identify all intercepts. (c) Find any vertical and horizontal asymptotes or holes. (d) Plot additional solution points as needed to sketch the graph of the rational function.

Consider the following. \[ \begin{array}{l}g(x)=\frac{5 x}{x^{2}+4 x-5} \\ \text { (a) State the domain of the function. } \\ \text { all real numbers } x \text { except } x=5 \\ \text { all real numbers } x \text { except } x=1 \\ \text { all real numbers } x \text { except } x=-5 \\ \text { all real numbers } x \text { except } x=-5 \text { and } x=1 \\ \text { (b) Identify all intercepts. (If an answer does not exist, enter DNE.) } \\ x \text { except } x=0 \\ x \text {-intercept } \quad(x, y)=(\square) \\ y \text {-intercept } \quad(x, y)=(\square)\end{array} \]