Exponential graphs can either decay or grow. This is based on the value of the base of the exponent. If the base is greater than $\displaystyle 1$, the graph will be growth. And, if the base is less than $\displaystyle 1$, then the graph will be decay. In this situation, our base is $\displaystyle 2$. Since this is greater than $\displaystyle 1$, we have a growth graph. Then, to determine the y-intercept we substitute $\displaystyle x=0$. Thus, we get: 

$\displaystyle y = 2^0 = 1$ for the y-intercept. 

Exponential graphs can either decay or grow. This is based on the value of the base of the exponent. If the base is greater than $\displaystyle 1$, the graph will be growth. And, if the base is less than $\displaystyle 1$, then the graph will be decay. In this situation, our base is $\displaystyle \frac{1}{3}$. Since this is less than $\displaystyle 1$, we have a decay graph. Then, to determine the y-intercept we substitute $\displaystyle x=0$. Thus, we get: 

$\displaystyle y = (\frac{1}{3})^{(4\cdot 0)} = (\frac{1}{3})^0=1$ for the y-intercept. 

Note that the negative sign in this function comes outside of the parentheses. This should show you that the bigger the number in parentheses, the lower the curve of the graph will go.  Since this is an exponential function, the larger that the x value gets, then, the "more negative" this graph will go. The graph closest to zero on the left-hand side - where x is negative - and then shoots down and to the right rapidly when x gets larger is the correct graph.

The $\displaystyle x$-coordinate ofthe $\displaystyle y$-intercept of the graph of $\displaystyle f$ is 0, and its $\displaystyle y$-coordinate is $\displaystyle f(0)$:

$\displaystyle f(x) = 4 ^{x+2} - 3$

$\displaystyle f(0) = 4 ^{0+2} - 3 = 4 ^{2} - 3 = 16 - 3 = 13$

The $\displaystyle y$-intercept is the point $\displaystyle (0,13)$.

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 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 $\displaystyle y$-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: 

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