The asymptote of this equation can be found by observing that $\displaystyle e^{x-3} > 0$ regardless of $\displaystyle x$. We are thus solving for the value of $\displaystyle y$ as $\displaystyle e^{x-3}$ approaches zero.

$\displaystyle -2e^{x-3} < -2\cdot 0$

$\displaystyle -2e^{x-3} < 0$

$\displaystyle -2e^{x-3} -5 < 0-5$

$\displaystyle -2e^{x-3} -5 < -5$

$\displaystyle y< -5$

So the value that $\displaystyle y$ cannot exceed is $\displaystyle -5$, and the line $\displaystyle y = -5$ is the asymptote.

An exponential equation of the form $\displaystyle f(x) = ae^{x-h} +k$ has only one asymptote - a horizontal one at $\displaystyle y = k$. In the given function, $\displaystyle k = -2$, so its one and only asymptote is $\displaystyle y = -2$.

To find the vertical asymptotes, we set the denominator of the function equal to zero and solve.

$\displaystyle 0=x^2-16$

$\displaystyle 16=x^2$

$\displaystyle \sqrt{16}=\sqrt{x^2}$

$\displaystyle 4, -4=x$

Factorize both the numerator and denominator.

$\displaystyle y=\frac{x^2-2x+1}{x^2-1}=\frac{(x-1)(x-1)}{(x+1)(x-1)}$

Notice that one of the binomials will cancel.

The domain of this equation cannot include $\displaystyle x=\pm1$.

The simplified equation is:

$\displaystyle y=\frac{x-1}{x+1}$

Since the $\displaystyle (x-1)$ term canceled, the $\displaystyle (x-1)$ term will have a hole instead of an asymptote.   

Set the denominator equal to zero.

$\displaystyle x+1=0$

Subtract one from both sides.

$\displaystyle x=-1$

There will be an asymptote at only:  $\displaystyle x=-1$

The answer is:  $\displaystyle x=-1$

Factor the numerator and denominator.

$\displaystyle y=\frac{2x^2+2x}{x^2-8x-9} = \frac{2x(x+1)}{(x-9)(x+1)}$

Notice that the $\displaystyle (x+1)$ terms will cancel.  The hole will be located at $\displaystyle x=-1$ because this is a removable discontinuity.

$\displaystyle y=\frac{2x}{x-9}$

The denominator cannot be equal to zero.  Set the denominator to find the location where the x-variable cannot exist.

$\displaystyle x-9\neq0$

$\displaystyle x\neq9$

The asymptote is located at $\displaystyle x=9$.

Factor the numerator and denominator.

$\displaystyle 2x-4 = 2(x-2)$

$\displaystyle x^2-4 = (x+2)(x-2)$

Rewrite the equation.

$\displaystyle y=\frac{2x-4}{x^2-4}=\frac{ 2(x-2)}{(x+2)(x-2)}=\frac{2}{x+2}$

Notice that the $\displaystyle (x-2)$ will cancel.  This means that the root of $\displaystyle (x-2)$ will be a hole instead of an asymptote.

Set the denominator equal to zero and solve for x.

$\displaystyle x+2=0$

$\displaystyle x=-2$

An asymptote is located at:  $\displaystyle x=-2$

The answer is:  $\displaystyle x=-2$

For positive $\displaystyle x$ values, $\displaystyle y$ increases exponentially in the $\displaystyle +x$ direction and goes to positive infinity, so there is no asymptote on the positive $\displaystyle x$-axis. For negative $\displaystyle x$ values, as $\displaystyle x$ decreases, the term $\displaystyle 2^x$ becomes closer and closer to zero so $\displaystyle y$ approaches $\displaystyle 3$ as we move along the negative $\displaystyle x$ axis. As the graph below shows, this is forms a horizontal asymptote.

Begin by recognizing that both sides of the equation have a root term of $\displaystyle 3$.

$\displaystyle \small 9^x=3^6$

$\displaystyle (3^2)^x=3^6$

Using the power rule, we can set the exponents equal to each other.

$\displaystyle 3^{(2*x)}=3^6$

$\displaystyle \small 2x=6$

$\displaystyle \small x=3$

Begin by recognizing that both sides of the equation have the same root term, $\displaystyle 3$.

$\displaystyle \small 3^{2x}=81$

$\displaystyle \small 3^{2x}=9^2$

$\displaystyle 3^{2x}=(3^2)^2$

We can use the power rule to combine exponents.

$\displaystyle 3^{2x}=3^4$

Set the exponents equal to each other.

$\displaystyle 2x=4$

$\displaystyle \small x=2$

Solve for the values of a and b:

In 2009, $\displaystyle y=1034$ and $\displaystyle x=0$ (zero years since 2009). Plug this into the exponential equation form:

$\displaystyle 1034=ab^{(0)}$. Solve for $\displaystyle a$ to get  $\displaystyle a=1034$.

In 2013, $\displaystyle y=1711$ and $\displaystyle x=4$. Therefore,

$\displaystyle 1711=ab^4$  or  $\displaystyle 1711=(1034)b^4$.   Solve for $\displaystyle b$ to get

$\displaystyle b=\sqrt[4]{\frac{1711}{1034}}\approx1.1342$.

Then the exponential growth function is  

$\displaystyle y=1034(1.1342)^{x}$.