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. 

Exponential functions with a base greater than one are models of exponential growth. Thus, we know that our function will increase and not decrease. Remembering the graph of an exponential function, we can determine that the graph will begin gradually, almost like a flat line. Then, as \(\displaystyle x\) increases, \(\displaystyle {}y\) begins to increase very quickly.