PreCalc: How to find an exponential function?

Question

This whole concept is confusing to me.
If the first point is (-2, 45/4) and the second point is (1, 10/3), how do I find the formula for exponential growth???

That doesn't exactly help me.

I know the exp version looks like ae^(kt) but I don't know how to get to that from just two points!

SOOOOOOOOOOOOOO
would the answer look like:
P(t) = 5e^(-.405465t)?

Answer 1

45/4 = Ae^(k*-2) and 10/3 = Ae^(k*1)
Solve for A-
A = (45/4) / [ e^(-2k) ] = (10/3) / [ e^(k) ]
Rearrange to get
(45/4) (e^(k)) = (10/3) (e^(-2k))
And again to get
e^(k) / (e^(-2k) = (10/3) / (45/4) = 8/27

Just as a^x / a^y = a^(x-y),
e^(k) / (e^(-2k) = e^(3k) = 8/27
Using logs, you SHOULD be able to solve for k.
Which you then plug back in to solve for A.

I didn't want to GIVE you the WHOLE thing :)

Answer 2

It is written as exp(x) or ex, where e equals approximately 2.71828183 and is the base of the natural logarithm.

In mathematics, exponential growth (or geometric growth) occurs when the growth rate of a function is always proportional to the function's current size. Such growth is said to follow an exponential law (but see also Malthusian growth model). This implies that for any exponentially growing quantity, the larger the quantity gets, the faster it grows. But it also implies that the relationship between the size of the dependent variable and its rate of growth is governed by a strict law, of the simplest kind: direct proportion. It is proved in calculus that this law requires that the quantity is given by the exponential function, if we use the correct time scale. This explains the name.
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We first start with the properties of the graph of the basic exponential function of base a,

f (x) = ax , a > 0 and a not equal to 1.

The domain of function f is the set of all real numbers. The range of f is the interval (0 , +infinity).

The graph of f has a horizontal asymptote given by y = 0. Function f has a y intercept at (0 , 1). f is an increasing function if a is greater than 1 and a decreasing function if a is smaller than 1 .

You may want to review all the above properties of the exponential function interactively .
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Example 1: f is a function given by
f (x) = 2(x - 2)

Find the domain and range of f.
Find the horizontal asymptote of the graph.
Find the x and y intercepts of the graph. of f if there are any.
Sketch the graph of f.

Answer to Example 1
The domain of f is the set of all real numbers. To find the range of f,we start with 2x > 0

Multiply both sides by 2-2 which is positive. 2x2-2 > 0

Use exponential properties 2(x - 2) > 0

This last statement suggests that f(x) > 0. The range of f is (0, +inf).

As x decreases without bound, f(x) = 2(x - 2) approaches 0. The graph of f has a horizontal asymptote at y = 0.

To find the x intercept we need to solve the equation f(x) = 0
2(x - 2) = 0

This equation does not have a solution, see range above, f(x) > 0. The graph of f does not have an x intercept. The y intercept is given by (0 , f(0)) = (0,2(0 - 2)) = (0 , 1/4).

So far we have the domain, range, y intercept and the horizontal asymptote. We need extra points.
(4 , f(4)) = (4, 2(4 - 2)) = (4 , 22) = (4 , 4)

(-1 , f(-2)) = (-1, 2(-1 - 2)) = (-1 , 2-3) = (-1 , 1/8)

Now use all the above information to graph f.
Matched Problem to Example1: f is a function given by
f (x) = 2(x + 2)

Find the domain and range of f.
Find the horizontal asymptote of the graph of f.
Find the x and y intercepts of the graph of f if there are any.
Sketch the graph of f.

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Example 2: f is a function given by f (x) = 3(x + 1) - 2

Find the domain and range of f.
Find the x and y intercepts of the graph of f if there are any.
Sketch the graph of f.

Answer to Example 2
The domain of f is the set of all real numbers. To find the range of f, we start with 3x > 0

Multiply both sides by 3 which is positive. 3x3 > 0

Use exponential properties 3(x + 1) > 0

Subtract 2 to both sides 3(x + 1) -2 > -2

This last statement suggests that f(x) > -2. The range of f is (-2, +inf).

As x decreases without bound, f(x) = 3(x + 1) -2 approaches -2. The graph of f has a horizontal asymptote y = -2.

To find the x intercept we need to solve the equation f(x) = 0
3(x + 1) - 2 = 0

Add 2 to both sides of the equation 3(x + 1) = 2

Rewrite the above equation in Logarithmic form x +1 = log3 2

Solve for x
x = log3 2 - 1

The y intercept is given by
(0 , f(0)) = (0,3(0 + 1) - 2) = (0 , 1).

So far we have the domain, range, x and y intercepts and the horizontal asymptote. We need extra points.
(-2 , f(-2)) = (-2, 3(-2 + 1) - 2) = (4 , 1/3-2) = (4 , -1.67)

(-4 , f(-4)) = (-4, 3(-4 + 1) - 2) = (-4 , 2-3) = (-4 , -1.99)

Now use all the above information to graph f.
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Matched Problem to Example2: f is a function given by
f (x) = 2(x - 2) + 1

Find the domain and range of f.
Find the horizontal asymptote of the graph of f.
Find the x and y intercepts of the graph of f if there are any.
Sketch the graph of f.

Answer 3

Hi Kale-- Growth/depletion usually refers to something that is a natural phenomenon and thus e usually appears in the equation but these are kinda like +/- interest rates on a current balance. Therefore, the increase is New Value = Old Value*(1+0.03)/yr. The other is a depletion equation which is New Value = Old Value*(1-0.01)/day. This way you can figure out the new values at any year or day from the equations above.