Answer:
B.[tex]f(x)=2(3^x)+4[/tex]
Step-by-step explanation:
We have to find that which graph has horizontal asymptote at 4
We know that to find the horizontal asymptote , we simply evaluate the limit of the function as it approaches to infinity or it approaches to negative infinity.
A.[tex]f(x)=2x-4[/tex]
[tex]\lim_{x\rightarrow \infty}(2x-4)=\infty[/tex]
[tex]\lim_{x\rightarrow -\infty}(2x-4)=-\infty[/tex]
Limit of function does not exits, so function have not horizontal asymptote.
B.[tex]f(x)=2(3^x)+4[/tex]
[tex]\lim_{x\rightarrow \infty}(2(3^x)+4)=\infty[/tex]
[tex]3^{\infty}=\infty [/tex]
[tex]\lim_{x\rightarrow -\infty}(2(3^x)+4)=4[/tex]
Because [tex]3^{-\infty}=0[/tex]
Hence, function have horizontal asymptote at 4.
C.[tex]f(x)=-3x+4[/tex]
[tex]\lim_{x\rightarrow \infty}(-3x+4)=\infty[/tex]
[tex]\lim_{x\rightarrow -\infty}(-3x+4)=-\infty[/tex]
Hence, function have not horizontal asymptote.
D.[tex]f(x)=3(2^x)-4[/tex]
[tex]\lim_{x\rightarrow \infty}(3(2^x)-4)=\infty[/tex]
Because [tex]2^{\infty}=\infty[/tex]
[tex]\lim_{x\rightarrow -\infty}(3(2^x)-4)=-4[/tex]
[tex]2^{-\infty}=0[/tex]
Hence, function have horizontal asymptote at -4.
Therefore, option B is true.
