# Find the exponential model that fits the points shown in the graph. (Round the exponent to four decimal places)

The aim of this question is to understand the exponential function, how to fit the points into the exponent model and understand what the exponential function describes.

In mathematics, the exponential function is described by a relation of the form y=a^x. where the independent variable x goes over the entire real number and a is a constant number that is greater than zero. a in exponential function is known as the base of the function. y=e^x or y=exp(x) is one of the most important exponential function where the e is 2.7182818, base of the natural system of logarithms (ln)

An exponential model grows or decays depending on the function. In exponential growth or exponential decay, an amount rises or falls by an appointed percent at regular intervals.

In exponential growth, the quantity rises slowly but increases rapidly after some intervals. As time passes, the rate of change becomes faster. This change in growth is marked as an exponential increase. The formula for exponential growth is denoted by:

$y = a(1+r)^x$

where $r$ represents the growth rate.

In exponential decay, The quantity falls rapidly at first but slows down after some intervals. As time passes, the rate of change becomes slower. This change in growth is marked as an exponential decrease. The formula for exponential decay is denoted by:

$y = a(1-r)^x$

where $r$ represents the decay percentage.

Given points are $(0,8)$ and $(1,3)$.

General equation of the exponential model is $y = ae^{bx}$.

So first we will take the point $(0,8)$ and substitute in the general equation and solve for $a$.

Inserting the $(0,8)$ in the general equation will eliminate $b$ as it will get multiplied by $0$ and hence will make it easy to solve for $a$:

$y = ae^{bx}$

Inserting $(0,8)$:

$8 =ae^{b(0)}$

$8 =ae^0$

Anything with power $0$ is $1$, so:

$a =8$

Now that the $a$ is known, Insert the point $(1,3)$ and solve for $b$:

$y=ae^{bx}$

$3=ae^{b(1)}$

Inserting $a=8$:

$3=8e^{b}$

$e^b=\dfrac{3}{8}$

Taking $ln$ to solve for $b$:

$b= ln(\dfrac{3}{8})$

Exponential model that fits the points $(0,8)$ and $(1,3)$ is $y = 8e^{ln \left(\dfrac{3}{8}\right) }$.

## Example

How do you find the exponential model $y=ae^{bx}$ that fits the two points $(0, 2)$, $(4, 3)$?

Given points are $(0,2)$ and $(4,3)$.

Exponential model in the question is given as $y = ae^{bx}$.

So first we will plug in the point $(0,8)$ in the general equation and solve for $a$.

Reason for plugging this point that by inserting $(0,8)$ in the given equation, it will eliminate $b$ and hence will make it easy to solve for $a$.

$y=ae^{bx}$

Inserting $(0,2)$:

$2=ae^{b(0)}$

$2=ae^0$

Anything with power $0$ is $1$ so:

$a =2$

Now that the $a$ is known, Insert the point $(4,3)$ and solve for $b$.

$y=ae^{bx}$

$3=ae^{b(4)}$

Inserting $a=2$:

$3= 2e^{4b}$

$e^{4b}= \dfrac{3}{2}$

Taking $ln$ to solve for $b$:

$4b= ln(\dfrac{3}{2})$

$b= \dfrac{ln(\dfrac{3}{2})}{4}$

Exponential model that fits the points $y=2e^{101x}$ $(0,2)$ and $(4,3)$ is $y = 2e^{0.101x}$.