91Ó°ÊÓ

The graph whose equation need to found out is as:

Since for each of the graph asymptote is given we will find the equation of asymptote and will match with the options available.

Considering the equations one by one we get:

$$\begin{gathered} \left(A\right) \\ y=3^x \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}{3}^x \\ =∞or0 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(B\right) \\ y=3^{-x} \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}{3}^{-x} \\ =0or∞ \\ whichisequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(C\right) \\ y=-3^x \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}-3^x \\ =∞or0 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(D\right) \\ y=-3^{-x} \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}-3^{-x} \\ =-∞or0 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(E\right) \\ y=3^x-1 \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}{3}^x-1 \\ =∞or-1 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(F\right) \\ y=3^{x-1} \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}{3}^{x-1} \\ =∞or0 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(G\right) \\ y=3^{1-x} \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}{3}^{1-x} \\ =∞or0 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(H\right) \\ y=1-3^x \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}1-3^x \\ =-∞or1 \\ whichisnotequaltoy=0 \end{gathered}$$

$$\begin{gathered} \left(H\right) \\ y=1-3^x \\ li{m}_{x→Â\pm ∞}y=li{m}_{x→Â\pm ∞}1-3^x \\ =-∞or1 \\ whichisnotequaltoy=0 \end{gathered}$$