Determine the equations of the asymptotes and all intercepts with the axes of the following exponential functions. In addition, determine whether the graph is increasing or decreasing and state whether the graph lies above or below the asymptote:

The general form of an exponential function is $y = a \cdot b^x + d$. Here, $a=1$, $b=2$, and $d=0$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = 0}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 1)}$.
For the x-intercept, set $y=0$:

There is no real value of $x$ for which $2^x = 0$.
There is $\boxed{\text{no x-intercept}}$.

Since the base $b=2 > 1$ and the coefficient $a=1 > 0$, the function is increasing.
The graph is $\boxed{\text{increasing}}$.

Since the coefficient $a=1 > 0$, the graph lies above the asymptote.
The graph lies $\boxed{\text{above the asymptote}}$.

Here, $a=1$, $b=2$, and $d=-1$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = -1}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 0)}$.
For the x-intercept, set $y=0$:

The x-intercept is $\boxed{(0, 0)}$.

Since the base $b=2 > 1$ and the coefficient $a=1 > 0$, the function is increasing.
The graph is $\boxed{\text{increasing}}$.

Since the coefficient $a=1 > 0$, the graph lies above the asymptote.
The graph lies $\boxed{\text{above the asymptote}}$.

Here, $a=1$, $b=2$, and $d=1$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = 1}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 2)}$.
For the x-intercept, set $y=0$:

There is no real value of $x$ for which $2^x = -1$.
There is $\boxed{\text{no x-intercept}}$.

Since the base $b=2 > 1$ and the coefficient $a=1 > 0$, the function is increasing.
The graph is $\boxed{\text{increasing}}$.

Since the coefficient $a=1 > 0$, the graph lies above the asymptote.
The graph lies $\boxed{\text{above the asymptote}}$.

The function can be written as $f(x) = -1 \cdot (2^{-1})^x + 4 = -1 \cdot \left(\frac{1}{2}\right)^x + 4$.
Here, $a=-1$, $b=\frac{1}{2}$, and $d=4$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = 4}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 3)}$.
For the x-intercept, set $f(x)=0$:

The x-intercept is $\boxed{(-2, 0)}$.

Since the coefficient $a=-1 < 0$ and the base $b=\frac{1}{2} < 1$, the function is increasing.
The graph is $\boxed{\text{increasing}}$.

Since the coefficient $a=-1 < 0$, the graph lies below the asymptote.
The graph lies $\boxed{\text{below the asymptote}}$.

Here, $a=-2$, $b=2$, and $d=2$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = 2}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 0)}$.
For the x-intercept, set $f(x)=0$:

The x-intercept is $\boxed{(0, 0)}$.

Since the coefficient $a=-2 < 0$ and the base $b=2 > 1$, the function is decreasing.
The graph is $\boxed{\text{decreasing}}$.

Since the coefficient $a=-2 < 0$, the graph lies below the asymptote.
The graph lies $\boxed{\text{below the asymptote}}$.

Here, $a=3$, $b=\frac{1}{3}$, and $d=2$. The horizontal asymptote is $y=d$.
The asymptote is $\boxed{y = 2}$.

For the y-intercept, set $x=0$:

The y-intercept is $\boxed{(0, 5)}$.
For the x-intercept, set $f(x)=0$:

There is no real value of $x$ for which $\left(\frac{1}{3}\right)^x$ is negative.
There is $\boxed{\text{no x-intercept}}$.

Since the coefficient $a=3 > 0$ and the base $b=\frac{1}{3} < 1$, the function is decreasing.
The graph is $\boxed{\text{decreasing}}$.

Since the coefficient $a=3 > 0$, the graph lies above the asymptote.
The graph lies $\boxed{\text{above the asymptote}}$.