sketch the graph of the following function
$f(x) = e^{-2x}$ for $x \in \mathbb R$
this what i got,
y-intercept $x=0$ implies $y=\cfrac{1}{e^{2\times 0}}$ therefore $y=1$
and I have used the fact that $t= -2x$ $\rightarrow -\infty$ as $x \rightarrow \infty$
by the properties of the Natural Exponential Function:
$\displaystyle\lim_{x \to \infty} e^x=\infty$
$\displaystyle\lim_{x \to \infty} e^{-2x}=0$
$\displaystyle\lim_{x \to -\infty} e^{t}=0$
Therefore, I draw the curve graph with $y$-intercept coordinates $(0,1)$ and approaches zero.
but my prof said it's wrong. he commented = $e^x$ cannot be zero.
Could you please explain to me about this?...
Your professor is correct (as usual) he describes asymptotic behaviour - namely $y=0$ is a horizontal asymptote for $e^{-2x}$. Sketch the graph of $e^{-2x}$ to see this. Firstly you know how to sketch the graph of $e^x$ right? You then apply linear transformations which consist of a horizontal reflection in the $y$-axis and then a stretch of scale factor $1/2$ parallel to the $x$-axis. Does this make sense? If not let me know and I will explain further.