An exponential function of the form 

$\displaystyle g(x) = a^{x- h}+k$

has as its one and only asymptote the horizontal line $\displaystyle y = k$. 

Since we define $\displaystyle g$ as 

$\displaystyle g(x) = 0.4 ^{x - 2} + 0.7$,

then $\displaystyle k= 0.7$, 

and the only asymptote is the line of the equation $\displaystyle y = 0.7$.

a)

This is exponential decay since the base, $\displaystyle (1/2)$, is between $\displaystyle 0$ and $\displaystyle 1$.

b)

This is exponential growth since the base, $\displaystyle 1.1$, is greater than $\displaystyle 1$.

For $\displaystyle y=2^x$, our base is greater than $\displaystyle 1$ so we have exponential growth, meaning the function is increasing. Also, when $\displaystyle x=0$, we know that $\displaystyle y=1$ since $\displaystyle 2^0=1$. The only graph that fits these conditions is $\displaystyle c$.

For $\displaystyle y=3(2^x)$, we have exponential growth again but when $\displaystyle x=0$, $\displaystyle y=3$. This is shown on graph $\displaystyle a$.

For $\displaystyle y=(1/2)^x$, we have exponential decay so the graph must be decreasing. Also, when $\displaystyle x=0$, $\displaystyle y=1$. This is shown on graph $\displaystyle b$.

To determine the constant $\displaystyle A$, we look at the graph to find the initial value of $\displaystyle y$ , (when $\displaystyle t=0$) and find it to be $\displaystyle y(0)= 1.2$.  We can then plug this into our equation $\displaystyle y=Ae^{kt}$ and we get $\displaystyle 1.2=Ae^{k0}$. Since $\displaystyle e^{k0}=e^0=1$, we find that $\displaystyle A=1.2$.

To find $\displaystyle k$, we use the fact that when $\displaystyle t=0.3648$, $\displaystyle y$ is one half of the initial value $\displaystyle y(0)=1.2$. Plugging this into our equation with $\displaystyle A$ now known gives us $\displaystyle \frac{1.2}{2}=0.6=1.2e^{k(0.3648)}$ . To solve for $\displaystyle k$, we make use the fact that the natural log is the inverse function of $\displaystyle e$, so that 

$\displaystyle \ln(e^{k(0.3648)})=k(0.3648)$.

We can write our equation as  $\displaystyle \frac{0.6}{1.2}=e^{k(0.3648)}$ and take the natural log of both sides to get:

$\displaystyle \ln(\frac{0.6}{1.2})=\ln(e^{k(0.3648)})$ or $\displaystyle -0.6931=k(0.3648)$.

Then $\displaystyle k\approx -1.9$.

Our model equation is $\displaystyle y=1.20e^{-1.9t}$.

We need to determine the constants $\displaystyle a$ and $\displaystyle b$. Since $\displaystyle y=416$ in 2010 (when $\displaystyle x=0$), then $\displaystyle y(0)=ab^0=a$ and $\displaystyle a=416$

To get $\displaystyle b$, we find that when $\displaystyle x=5$, $\displaystyle y(5)=416b^5=521$.  Then  $\displaystyle b^5=\frac{521}{416}$.

Using a calculator, $\displaystyle {\left(\frac{521}{416}\right)}^{\frac{1}{5}}=1.046$, so $\displaystyle b\approx 1.046$.

Then our model equation for the fish population is $\displaystyle y=416(1.046)^x$

The $\displaystyle y$-intercept of any graph describes the $\displaystyle y$-value of the point on the graph with a $\displaystyle x$-value of $\displaystyle 0$.

Thus, to find the $\displaystyle y$-intercept substitute $\displaystyle x = 0$.

In this case, you will get,

 $\displaystyle y = 3^{0} = 1$. 

The $\displaystyle y$-intercept of a graph is the point on the graph where the $\displaystyle x$-value is $\displaystyle 0$.

Thus, to find the $\displaystyle y$-intercept, substitute $\displaystyle x = 0$ and solve for $\displaystyle y$.

Thus, we get: 

$\displaystyle y = 4(2^{0}) = 4(1)=4$

The $\displaystyle y$-intercept of any function describes the point where $\displaystyle x = 0$.

Substituting this in to our funciton, we get: 

$\displaystyle y = 5^{x+2} = 5^{0+2} = 5^{2} = 25$

Exponential decay describes a function that decreases by a factor every time $\displaystyle x$ increases by $\displaystyle 1$.

These can be recognizable by those functions with a base which is between $\displaystyle 0$ and $\displaystyle 1$.

The general equation for exponential decay is,

$\displaystyle A=A_0b^t$ where the base is represented by $\displaystyle b$ and $\displaystyle 0< b< 1$.

Thus, we are looking for a fractional base.

The only function that has a fractional base is,

 $\displaystyle y = \left(\frac{1}{3}\right)^{x}$. 

The $\displaystyle x$-intercept of a function is where $\displaystyle y = 0$. Thus, we are looking for the $\displaystyle x$-value which makes $\displaystyle 0 = 4^{x}$.

If we try to solve this equation for $\displaystyle x$ we get an error.

To bring the exponent down we will need to take the natural log of both sides.

$\displaystyle ln(0)=xln(4)$

Since the natural log of zero does not exist, there is no exponent which makes this equation true.

Thus, there is no $\displaystyle x$-intercept for this function.