All of the functions are of the form . To find the -intercept of such a function, we can set  and solve for :

Since we are looking for a function whose graph has -intercept , the equation here becomes , and we can examine each of the functions by finding the value of .

:

: 

: 

: 

All four choices fit the criterion.

All of the functions take the form 

for some integer . To find the choice that has -intercept , set  and , and solve for :

In exponential form:

The correct choice is .

Set  and evaluate  to find the -coordinate of the -intercept.

Rewrite in exponential form:

.

The -intercept is .

Since , the definition of  can be rewritten as follows:

Since , the definition of  can be rewritten as follows:

First, we need to find the -coordinate of the point at which the graphs of  and  meet by setting 

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

However, if we evaluate , the expression becomes

,

which is undefined, since a negative number cannot have a logarithm.

Consequently, the two graphs do not intersect.

All of the functions are of the form . To find the -intercept of a function ,  we can set  and solve for :

.

Since we are looking for a function whose graph has -intercept , the equation here becomes , and we can examine each of the functions by finding the value of  and seeing which case yields this result.

:

:

:

:

The graph of  has -intercept  and is the correct choice.

The -intercept of the graph of  can befound by setting  and solving for :

Rewritten in exponential form:

The -intercept of the graph of  is .

The -intercept of the graph of  can be found by evaluating 

The -intercept of the graph of  is .

If  and  are the - and -intercepts, respectively, of a line, the slope of the line is . Substituting  and , this is

.

Setting  and  in the slope-intercept form of the equation of a line:

Since , the definition of  can be rewritten as follows:

 . 

Find the -coordinate of the point at which the graphs of  and  meet by setting 

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

The quadradic trinomial can be "reverse-FOILed" by noting that 2 and 6 have  product 12 and sum 8:

Either , in which case 

or

, in which case 

Note, however, that we can eliminate  as a possible -value, since

,

an undefined quantity since negative numbers do not have logarithms. 

Since 

and 

,

 is the correct -value, and  is the correct -value.

The -coordinate(s) of the -intercept(s) are the real solution(s) to the equation . We can use the quadratic formula to find any solutions, setting  - the coefficients of the expression.

An examination of the discriminant , however, proves this unnecessary.

The discriminant being negative, there are no real solutions, so the parabola has no  -intercepts.

When plotting a complex number, we use a set of real-imaginary axes in which the x-axis is represented by the real component of the complex number, and the y-axis is represented by the imaginary component of the complex number. The real component is    and the imaginary component is  ,  so this is the equivalent of plotting the point    on a set of Cartesian axes.  Plotting the complex number on a set of real-imaginary axes, we move    to the left in the x-direction and    up in the y-direction, which puts us in the second quadrant, or in terms of Roman numerals:

If we graphed the given complex number on a set of real-imaginary axes, we would plot the real value of the complex number as the x coordinate, and the imaginary value of the complex number as the y coordinate. Because the given complex number is as follows:

We are essentially doing the same as plotting the point    on a set of Cartesian axes.  We move    units right in the x direction, and    units down in the y direction, which puts us in the fourth quadrant, or in terms of Roman numerals: