Exponential functions and graphs

Summary and examples

The following formula is an example of an exponential function:

y=a^x

where a>0. The name exponential function is used if the exponent contains the independent variable x. The function above is always positive because the power of a positive number is always positive.
This function can be extended, for example:

y=Aa^{cx+d}+e

A, a, c, d and e are constants and are sometimes called parameters.

We will draw the graphs of two exponential functions.
The left graph corresponds to the function:

y=2^x

and the right graph corresponds to the function:

y=(\displaystyle\frac{1}{2})^x

 2^x   (0.5)^x

Both graphs are each other’s mirror, they are mirrored in the Y-axis. We can easily show this algebraically:

y=(\displaystyle\frac{1}{2})^x=(2^{-1})^x=2^{-x}

and this is the mirror in the Y-axis of the function:

y=2^x

We will bring together in one figure some functions of the type :

y=a^x

namely:

y=1.5^x

y=2^x

y=3^x

See the figure.

(1.5)^x en 2^x en 3^x

We first notice that all graphs go through the point (0,1). This is because we have for x=0:

y=a^0=1

for each a>0.

Furthermore we can easily explain that the fastest increasing graph (green) corresponds to the function:

y=3^x

and the least fast increasing graph (blue) to the function:

y=1.5^x

We also see that all three graphs converge to 0 when x will be more and more negative. In other words, if:

x \rightarrow-\infty then y\rightarrow0

The graphs will not reach but only approach the X-axis. Such a line, in this case the X-axis, is called an asymptote.

Now we look at a somewhat more complex exponential function:

y=3\cdot2^{2x-4}-1

In the figure the graph of this function is outlined.

3.2^(2x-4)-1

If x=0 we get:

y=3\cdot2^{-4}-1=\displaystyle\frac{3}{16}-1=-\displaystyle\frac{13}{16}

and the figure shows that this seems to be correct.
If we want to know where this function intersects the X-axis, then we have to take y=0. Then we get the following exponential equation (see also Exponential equations):

3\cdot2^{2x-4}-1=0

2^{2x-4}=\displaystyle\frac{1}{3}

When we apply the logarithmic function to both the left- and right-hand side (see also Logarithmic functions and graphs), we get:

(2x-4)\log{(2)}=\log{(\displaystyle\frac{1}{3})}=\log{(3^{-1})}=-\log{3}

2x-4=-\displaystyle\frac{\log{(3)}}{\log{(2)}}

x=2-\displaystyle\frac{\log{(3)}}{2\log{(2)}}\approx1.2144

and we can see this in the figure. (Often it is not necessary to calculate an exact number and it is sufficient to get a result with an logarithm).
Finally we see that if x will get more and more negative, so if x\rightarrow-\infty the first term of the function approaches 0 and the graph approaches the value -1. We can see this in the figure.

Example 1

Draw the graph of the function:

y=4^x

We try to find a number of points which enables us to draw the graph.

For x=-2, -1, 0, 1, 2, we get respectively:

y=4^{-2}=\displaystyle\frac{1}{16}

y=4^{-1}=\displaystyle\frac{1}{4}

y=4^0=1

y=4^1=4

y=4^2=16

and thus we can draw the graph:

4^x

Voorbeeld 2

Draw the graph of the function:

y=3^x+1

We consider the function:

y=3^x

Next, we can draw the original function by adding 1 to the result.
We try to find a number of points which help us to draw the graph.
For x=-2, -1, 0, 1, 2 we get respectively:

y=3^{-2}=\displaystyle\frac{1}{9}

y=3^{-1}=\displaystyle\frac{1}{3}

y=3^0=1

y=3^1=3

y=3^2=9

See the figure.

3^x

We get the final result by adding 1 to this graph:

3^x+1

Example 3

Draw the graph of the functions:

f: y=2^x

g: y=(\displaystyle\frac{1}{2})^{x-1}-1

and calculate the intersection point.

We roughly know the graphs, see above. In the figure below they are drawn exactly. Verify which graph corresponds to which function.

2^x en (0,5)^(x-1)-1

In order to calculate the intersection point of both graphs we have to solve the equation:

2^x=(\displaystyle\frac{1}{2})^{x-1}-1.

We have to work out the equation (see also Exponential equations):

2^x=(\displaystyle\frac{1}{2})^{x-1}-1

2^x=2^{-x+1}-1

2^{2x}-2+2^x=0

2^{2x}+2^x-2=0

To derive the third equation from the second, both sides are multiplied by 2^x, by which we get a quadratic equation in the variable 2^x, see also Quadratic equations (abc-formula) and Quadratic equations (factorizing).
Now we get an alternative equation by taking:

z=2^x

and get:

z^2+z-2=0

We can factorize this equation:

z^2+z-2=(z+2)(z-1)=0

which has the solution:

z=-2 of z=1

The first solution is no real solution for x because an exponential function cannot be negative. The other solution yields:

2^x=1

x=0

Thus both graphs intersect in x=0, which could be expected from the figure.

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