As x tends to negative infinity, the expression is asymptotically 0.
To determine if ( y = x ) is a horizontal asymptote for the function ( y = e^x ), we need to analyze the behavior of ( e^x ) as ( x ) approaches infinity. As ( x ) increases, ( e^x ) grows exponentially and does not approach ( x ). Therefore, ( y = x ) is not a horizontal asymptote for ( y = e^x ). In fact, ( e^x ) has no horizontal asymptotes.
It is y = 0
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
if you mean e to the x power times log of x, it is e to the x divided by x
Yes, the asymptote is x = 0. In order for logarithmic equation to have an asymptote, the value inside log must be 0. Then, 5x = 0 → x = 0.
Answer: no [but open to debate] ((x-1)(x-2)(x+2))/(x-3) (x^2-3x+2)/(x-2)(x+2) Asymptote missing, graph it, there is no Asymptote because the (x-2)(x+2) can be factored out. yes
The graph of an exponential function f(x) = bx approaches, but does not cross the x-axis. The x-axis is a horizontal asymptote.
-1
integral of e to the power -x is -e to the power -x
what symbol best describes the asymptote of an exponential function of the form F(x)=bx
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0