How to graph an exponential function - Geometry

We can rewrite the statements using for both and as follows:

To solve this, we can set the expressions equal, as follows:

We can rewrite the statements using for both and as follows:

To solve this, we can multiply the first equation by , then add:

The -intercept is , where :

The -intercept is .

The -intercept is :

is the -intercept.

We can rewrite this as follows:

This is a translation of the graph of , which has as its horizontal asymptote, to the right two units and down three units. Because of the latter translation, the horizontal asymptote is .

Since 4 can be raised to the power of any real number, the domain of is the set of all real numbers. Therefore, there is no vertical asymptote of the graph of .

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate is :

,

The -intercept is the point .

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate can be found by setting equal to 0 and solving for . Therefore, we need to find such that . However, any power of a positive number must be positive, so for all real , and has no real solution. The graph of therefore has no -intercept.

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, and for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

Since any number, positive or negative, can appear as an exponent, the domain of the function is the set of all real numbers; in other words, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

The -coordinate ofthe -intercept of the graph of is 0, and its -coordinate is :

The -intercept is the point .

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate can be found by setting equal to 0 and solving for . Therefore, we need to find such that

.

The -intercept is therefore .

First, we rewrite both functions with a common base:

is left as it is.

can be rewritten as

To find the point of intersection of the graphs of the functions, set

The powers are equal and the bases are equal, so we can set the exponents equal to each other and solve:

To find the -coordinate, substitute 4 for in either definition:

, the correct response.

First, we rewrite both functions with a common base:

is left as it is.

can be rewritten as

To find the point of intersection of the graphs of the functions, set

Since the powers of the same base are equal, we can set the exponents equal:

Now substitute in either function:

, the correct answer.

Rewrite the system as

and substitute and for and , respectively, to form the system

Add both sides:

.

Now backsolve:

Now substitute back:

and

.

Define . In terms of , can be restated as

The graph of is a transformation of that of . As an exponential function, has a graph that has no vertical asymptote, as is defined for all real values of ; it follows that being a transformation of this function, also has a graph with no vertical asymptote as well.

Define . In terms of , can be restated as

.

The graph of has as its horizontal asymptote the line of the equation . The graph of is a transformation of that of - a right shift of 3 units ( ), a vertical stretch ( ), and a downward shift of 7 units ( ). The right shift and the vertical stretch do not affect the position of the horizontal asymptote, but the downward shift moves the asymptote to the line of the equation . This is the correct response.