Functions as Graphs - Algebra 2

The vertical line test can be used to determine if an equation is a function. In order to be a function, there must only be one (or ) value for each value of . The vertical line test determines how many (or ) values are present for each value of . If a single vertical line passes through the graph of an equation more than once, it is not a function. If it passes through exactly once or not at all, then the equation is a function.

The horizontal line test can be used to determine if a function is one-to-one, that is, if only one value exists for each (or ) value. Calculating zeroes, domain, and range can be useful for graphing an equation, but they do not tell if it is a function.

Example of a function:

Example of an equation that is not a function:

As is clear from the graph, in the interval between ( included) to , the is constant at and then from ( not included) to ( not included), the is a decreasing function.

Perpendicular lines have slopes that are negative reciprocals. This is the case with these two lines. Although these lines interesect, this is not the most appropriate answer since it does not account for the fact that they are perpendicular.

To find the slope of an equation first get the equation in slope intercept form.

where,

represents the slope.

Thus

When determining how a the graph of a function will be translated, we know that anything that happens to x in the function will impact the graph horizontally, opposite of what is expressed in the function, whereas anything that is outside the function will impact the graph vertically the same as it is in the function notation.

For this graph:

The graph will move 3 spaces left, because that is the opposite sign of the what is connected to x directly.

Also, the graph will move down 2 spaces, because that is what is outside the function and the 2 is negative.

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator and denominator.

The numerator is a perfect square trinomial and can be factored as such:

The denominator can be factored as the difference of squares:

Rewrite

as

The expression can be reduced by cancelling in both halves:

Set the denominator equal to 0 and solve:

The only asymptote is therefore the line of the equation .

To identify a function as even odd, or neither, determine by replacing with , then simplifying. If , the function is even; if is odd.

,

so

By the Power of a Product Property,

,

so is an odd function

To identify a function as even odd, or neither, determine by replacing with , then simplifying. If , the function is even; if is odd.

so

By the Power of a Product Property,

, so is not an even function.

,

, so is not an odd function.

To identify a function as even, odd, or neither, determine by replacing with , then simplifying. If , the function is even; if is odd.

Since , is an odd function.

To identify a function as even, odd, or neither, determine by replacing with , then simplifying. If , the function is even; if is odd.

, so is an even function.

A function is odd if and only if, for every in its domain, ; it is even if and only if, for every in its domain, . We can see that

Of course,

.

Therefore, is even by definition.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as seen below:

Also, it is seen to be symmetrical about the -axis. This proves the function even.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it can be seen to be symmetrical about the origin. Consequently, for each in the domain, - the function is odd.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it is seen to be symmetrical about the -axis. Consequently, for each in the domain, - the function is even.

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for :

The graph of has the line of the equation as its only vertical asymptote.

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator. It is a quadratic trinomial with lead term , so look to "reverse-FOIL" it as

by finding two integers with sum and product 30. By trial and error, these integers can be found to be and , so

Therefore, can be rewritten as

.

Cancelling , this can be seen to be essentially a polynomial function:

,

which does not have a vertical asymptote.

is a rational function whose numerator is a polynomial with degree greater than that of the polynomial in its denominator (2 and 1, respectively). The graph of such a function does not have a horizontal asymptote.

is a rational function whose numerator and denominator have the same degree (1). As such, it has as a horizontal asymptote the line of the equation , where is the quotient of the coefficients of the highest-degree terms of its numerator and denominator. Consequently, the horizontal asymptote of

is

,

or

.

is a rational function whose numerator and denominator are polynomials of the same degree (1). As such, it has as a horizontal asymptote the line of the equation , where is the quotient of the coefficients of the highest-degree terms of its numerator and denominator. Consequently, the horizontal asymptote of

is

or

.