Graphing Rational Functions Lesson

We have previously studied rational expressions. We learned that a rational expression is an algebraic fraction that is the quotient of two polynomials with a non-zero denominator. A rational function is a function that is defined by a rational expression. $$f(x) = \frac{p(x)}{q(x)}, q(x) ≠ 0$$ Where p(x) and q(x) are polynomials and the denominator q(x) is not zero.

Domain of a Rational Function

We previously learned how to find the domain of a rational function. Since we can't divide by zero, we set the denominator of our rational expression equal to zero and solve the resulting equation. Once we have the solutions to the equation, those values must be excluded from the domain. For reasons we will explain later on in the tutorial, we should always find the domain of a rational function before doing any simplifying. Any restrictions in the domain need to be carried over to the simplified version of the function.
Example #1: Find the domain of the rational function. $$f(x) = \frac{x - 1}{5x^2 - 34x - 7}$$ Here, we will set the denominator equal to zero and solve the resulting equation. $$5x^2 - 34x - 7 = 0$$ Factor the left side: $$(5x + 1)(x - 7) = 0$$ Zero-Product Property: $$5x + 1 = 0$$ $$\text{or}$$ $$x - 7 = 0$$ Let's start with the top equation: $$5x + 1 = 0$$ $$5x = -1$$ $$x = -\frac{1}{5}$$ Now let's go to the bottom equation: $$x - 7 = 0$$ $$x = 7$$ The solutions for our equation would be: $$x = -\frac{1}{5}, 7$$ Since these values create a denominator that is zero, they need to be restricted from the domain of our rational function. $$f(x) = \frac{x - 1}{5x^2 - 34x - 7}$$ $$\text{domain:} \, \left\{x | x ≠ -\frac{1}{5}, 7\right\}$$

Understanding Arrow Notation

It is important to understand arrow notation as it will be used throughout this tutorial.

Graphing y = 1/x

The most basic rational function with a variable in the denominator is y = 1/x or f(x) = 1/x if using function notation. This function is known as the reciprocal function, which we studied earlier in the course. The reciprocal function pairs every non-zero real number with its reciprocal. Since x is in the denominator, x can't be zero. $$\text{Domain:} \, \left\{x | x ≠ 0\right\}$$ $$\text{Range:} \, \left\{y | y ≠ 0\right\}$$ Desmos Link for More Detail

Features of f(x) = 1/x:

• Domain: {x | x ≠ 0}
• Range: {y | y ≠ 0}
• f(x) = 1/x decreases on the intervals:
  • (-∞, 0) and (0, ∞)
• The graph is discontinuous at x = 0
• The line x = 0 or the y-axis is a vertical asymptote
• The line y = 0 or the x-axis is a horizontal asymptote
• The reciprocal function f(x) = 1/x is an odd function
  • The graph is symmetric with respect to the origin
• f(x) → -∞ as x → 0^-
  • y approaches negative infinity as x approaches 0 from the left
• f(x) → ∞ as x → 0⁺
  • y approaches positive infinity as x approaches 0 from the right
• f(x) → 0 as x → -∞
  • y approaches 0 as x approaches negative infinity
• f(x) → 0 as x → ∞
  • y approaches 0 as x approaches positive infinity

Thinking carefully about the final four items on the list will allow us to see why we have a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. Keep in mind that if you divide 1 by a really large positive number, the result will be a number that is very close to 0. You can get as close to 0 as you would like by increasing the size of the divisor. Additionally, if you divide 1 by a really small positive number, one that is very close to zero, you will get a very large positive number. You can create a number as large as you would like by decreasing the size of the positive divisor, in other words, moving the divisor closer to zero. Let's begin with the first two items on the list. $$f(x) → -∞ \, \text{as} \, x → 0^-$$ This tells us that our y-values or function values f(x) will approach negative infinity as x approaches 0 from the left. Let's think about this with a table.

x	f(x)
-0.1	-10
-0.01	-100
...	...
-0.0000001	-10,000,000

$$f(x) → ∞ \, \text{as} \, x → 0^+$$ This tells us that our y-values or function values f(x) will approach positive infinity as x approaches 0 from the right. Let's think about this with a table.

x	f(x)
0.1	10
0.01	100
...	...
0.0000001	10,000,000

We can write our two scenarios as one using absolute value bars: $$|f(x)| → ∞ \, \text{as} \, x → 0$$ As x gets closer to 0 coming from either the left or the right, the y-values in terms of absolute value get larger and larger. Since x can't actually be equal to zero our graph will never intersect the vertical line x = 0. As we have previously seen, this line is known as a vertical asymptote.
Let's now consider the final two items on the list. As we have just seen, we can write the two scenarios as one using absolute value bars: $$f(x) → 0 \, \text{as} \, |x| → ∞$$ From the table above, we can see that as x gets large in terms of absolute value (|x| → ∞), the y-values get close to zero. In other words, the graph will get closer and closer to the horizontal line y = 0, which is known as a horizontal asymptote.

Graphing Transformations of y = 1/x

A simple rational function of the form: $$g(x) = \frac{ax + b}{cx + d}$$ can be graphed using the techniques of graphing transformations that we previously studied. Let's look at some examples.
Example #2: Graph each rational function. $$g(x) = \frac{2}{x - 1}$$ First, let's think about the parent function. $$f(x) = \frac{1}{x}$$ Rewrite g(x) in terms of f(x): $$g(x) = 2f(x - 1)$$ Based on the graph of f(x), g(x) will be vertically stretched by a factor of 2 and shifted 1 unit right. Since there aren't any vertical shifts, there will still be a horizontal asymptote at y = 0. Since there is a horizontal shift (1 unit right), the vertical asymptote will now occur at x = 1. So for a given (x, y) on the graph of f(x), the point will be (x + 1, 2y) on the graph of g(x). Desmos Link for More Detail Example #3: Graph each rational function. $$g(x) = \frac{2x + 5}{x + 1}$$ At first glance, it may not appear as though we can use graphing transformations to graph g(x) based on the parent function f(x) = 1/x. Notice that both the numerator and denominator have a degree of one. Let's use a simple long division to write our function in a different form. $$g(x) = (2x + 5) \, ÷ \, (x - (-1))$$ Since the divisor is of the form x - k, we can use synthetic division. Using the results of our synthetic division, we can rewrite g(x) as: $$g(x) = \frac{2x + 5}{x + 1} = 2 + \frac{3}{x + 1}$$ Now it is clear that g(x) can be graphed using transformations based on the parent function f(x) = 1/x. $$f(x) = \frac{1}{x}$$ $$g(x) = \frac{3}{x + 1} + 2$$ Rewrite g(x) in terms of f(x): $$g(x) = 3f(x + 1) + 2$$ Based on the graph of f(x), g(x) will be vertically stretched by a factor of 3, shifted 1 unit left, and shifted 2 units up. Since there is a vertical shift (up 2 units), the horizontal asymptote will now occur at y = 2. Additionally, since there is a horizontal shift (1 unit left), the vertical asymptote will now occur at x = -1. So for a given (x, y) on the graph of f(x), the point will be (x - 1, 3y + 2) on the graph of g(x). Desmos Link for More Detail

Graphing y = 1/x² and Transformations

Another very common rational function is y = 1/x² or f(x) = 1/x² if using function notation. f(x) = 1/x² is even, meaning it has y-axis symmetry. Similar to the reciprocal function, the graph has a break and is made up of two distinct branches. $$\text{Domain:} \, \left\{x | x ≠ 0\right\}$$ $$\text{Range:} \, \left\{y | y > 0\right\}$$ Desmos Link for More Detail

Features of y = 1/x²:

• Domain: {x | x ≠ 0}
• Range: {y | y > 0}
• f(x) = 1/x² increases on the interval (-∞, 0)
• f(x) = 1/x² decreases on the interval (0, ∞)
• The graph is discontinuous at x = 0
• The line x = 0 or the y-axis is a vertical asymptote
• The line y = 0 or the x-axis is a horizontal asymptote
• f(x) = 1/x² is an even function
  • The graph is symmetric with respect to the y-axis

Example #4: Graph each rational function. $$g(x) = \frac{2}{\left(\frac{1}{2}x + 3\right)^2} - 4$$ First, let's think about the parent function. $$f(x) = \frac{1}{x^2}$$ Rewrite g(x) in terms of f(x): $$g(x) = 2f\left[\frac{1}{2}(x + 6)\right] - 4$$ Based on the graph of f(x), g(x) will be vertically stretched by a factor of 2, horizontally stretched by a factor of 2, shifted 6 units left, and shifted 4 units down. Since there is a vertical shift (4 units down), the horizontal asymptote will now occur at y = -4. Since there is a horizontal shift (6 units left), the vertical asymptote will now occur at x = -6. So for a given (x, y) on the graph of f(x), the point will be (2x - 6, 2y - 4).

x	f(x)	x	g(x)
±2	1/4	-10, -2	-7/2
±1	1	-8, -4	-2
±1/2	4	-7, -5	4
0	undefined	-6	undefined

Desmos Link for More Detail

Vertical and Horizontal Asymptotes

We have seen with f(x) = 1/x, that we have a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. The line x = a is a vertical asymptote of the graph of a function if the function values increase or decrease without bound (approaches positive or negative infinity) as x approaches a from the right or the left. The image above shows an example of how a vertical asymptote might look if y approaches positive infinity as x approaches a from either the right or the left. The image above shows an example of how a vertical asymptote might look if y approaches negative infinity as x approaches a from either the right or the left.
The line y = b is a horizontal asymptote of the graph of a function if y approaches b as x increases or decreases without bound. The image above shows an example of how a horizontal asymptote might look if y approaches b as x approaches positive infinity (leftmost image) or negative infinity (rightmost image).

Features of Vertical and Horizontal Asymptotes

$$f(x) = \frac{p(x)}{q(x)}, q(x) ≠ 0$$ We will say the above is written in lowest terms, where the following a and b represent real numbers.

• If |f(x)| → ∞ as x → a, then x = a is a vertical asymptote
  • A rational function can have zero, one, or more than one vertical asymptotes
  • The graph of a rational function can never intersect a vertical asymptote
• If f(x) → b as |x| → ∞, then y = b is a horizontal asymptote
  • A rational function can have at most one horizontal asymptote
  • The graph of a rational function can intersect a horizontal asymptote

Locating Asymptotes

We can find the vertical asymptotes by finding the values for x that make the denominator 0. For horizontal asymptotes, we need to think about what happens to our y-values as x approaches positive or negative infinity. $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1}+ \cdots + a_1x + a_0}{b_mx^m + b_{m - 1}x^{m - 1}+ \cdots + b_1x + b_0}$$ Where: $$a_n, b_m ≠ 0$$ n is the degree of the numerator and m is the degree of the denominator.

• Vertical Asymptotes:
  • If a is a zero of the denominator, then the line x = a is a vertical asymptote
• Horizontal Asymptotes:
  • If n < m, then there is a horizontal asymptote y = 0 or the x-axis
    • In other words, if the numerator has a lower degree than the denominator
  • If n = m, then there is a horizontal asymptote y = a_n/b_m
    • In other words, if the numerator and denominator have the same degree:
      • y = the ratio of the leading coefficients is the horizontal asymptote
  • If n > m, then there won't be a horizontal asymptote
    • In other words, if the numerator has a greater degree than the denominator

Let's look at a few examples.
Example #5: Find all asymptotes. $$f(x) = \frac{x - 5}{3x^2 - 7x + 2}$$ To find the vertical asymptotes, set the denominator equal to zero and solve the resulting equation. $$3x^2 - 7x + 2 = 0$$ Factor the left side: $$(3x - 1)(x - 2) = 0$$ Zero-Product Property: $$3x - 1 = 0$$ $$\text{or}$$ $$x - 2 = 0$$ Top Equation: $$3x - 1 = 0$$ $$3x = 1$$ $$x = \frac{1}{3}$$ Bottom Equation: $$x - 2 = 0$$ $$x = 2$$ The equations of the vertical asymptotes are: $$x = \frac{1}{3} \, \text{and} \, x = 2$$ Since the degree of the numerator (1) is less than the degree of the denominator (2), we know that we have a horizontal asymptote y = 0 or the x-axis. We can also consider where this comes from by thinking about what happens to the y-values as x increases or decreases without bound. $$f(x) = \frac{x - 5}{3x^2 - 7x + 2}$$ Divide both numerator and denominator by x², which is the highest power of x in the expression. $$f(x) = \frac{x - 5}{3x^2 - 7x + 2} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$ $$= \large{\frac{\frac{x}{x^2} - \frac{5}{x^2}}{\frac{3x^2}{x^2} - \frac{7x}{x^2} + \frac{2}{x^2}}}$$ $$=\large{\frac{\frac{1}{x} - \frac{5}{x^2}}{3 - \frac{7}{x} + \frac{2}{x^2}}}$$ As |x| approaches infinity, the quotients 1/x, 5/x², 7/x, and 2/x² all approach 0, this means the value of f(x) will approach 0. $$\frac{0 - 0}{3 - 0 + 0} = \frac{0}{3} = 0$$ Our analysis above confirms that the equation of our horizontal asymptote is y = 0 or the x-axis. Desmos Link for More Detail Example #6: Find all asymptotes. $$f(x) = \frac{2x - 9}{x - 4}$$ To find the vertical asymptotes, set the denominator equal to zero and solve the resulting equation. $$x - 4 = 0$$ $$x = 4$$ The equation of the vertical asymptote is: $$x = 4$$ Since the degree of the numerator (1) is the same as the degree of the denominator (1), we know that we have a horizontal asymptote y = 2. This comes from taking the leading coefficient in the numerator (2) and placing it over the leading coefficient in the denominator (1). We can also consider where this comes from by thinking about what happens to the y-values as x increases or decreases without bound. $$f(x) = \frac{2x - 9}{x - 4}$$ Divide both numerator and denominator by x, which is the highest power of x in the expression. $$f(x) = \frac{2x - 9}{x - 4} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}$$ $$=\large{\frac{\frac{2x}{x} - \frac{9}{x}}{\frac{x}{x} - \frac{4}{x}}=\frac{2 - \frac{9}{x}}{1 - \frac{4}{x}}}$$ As |x| approaches infinity, the quotients 9/x and 4/x approach 0, this means the value of f(x) will approach 2/1, which is 2. $$\frac{2 - 0}{1 - 0} = \frac{2}{1} = 2$$ Our analysis above confirms that the equation of our horizontal asymptote is y = 2. Desmos Link for More Detail Example #7: Find all asymptotes. $$f(x) = \frac{x}{x^2 + 1}$$ To find the vertical asymptotes, set the denominator equal to zero and solve the resulting equation. $$x^2 + 1 = 0$$ $$x^2 = -1$$ $$x = \pm i$$ Since there are no real number solutions to the equation, we can say the domain is all real numbers. This means we will not have any vertical asymptotes.
Since the degree of the numerator (1) is less than the degree of the denominator (2), we know that we have a horizontal asymptote y = 0 or the x-axis. We can also consider where this comes from by thinking about what happens to the y-values as x increases or decreases without bound. Divide both numerator and denominator by x², which is the highest power of x in the expression. $$f(x) = \frac{x}{x^2 + 1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$ $$= \large{\frac{\frac{x}{x^2}}{\frac{x^2}{x^2} + \frac{1}{x^2}}}$$ $$= \large{\frac{\frac{1}{x}}{1 + \frac{1}{x^2}}}$$ As |x| approaches infinity, the quotients 1/x and 1/x² approach 0, this means the value of f(x) will approach 0. $$\frac{0}{1 + 0} = \frac{0}{1} = 0$$ Our analysis above confirms that the equation of our horizontal asymptote is y = 0 or the x-axis.
Note: In this case, the y-value can be 0. $$f(0) = \frac{0}{0^2 + 1} = 0$$ This demonstrates that a graph may cross a horizontal asymptote. Desmos Link for More Detail From the graph, notice that y approaches 0 as x increases or decreases without bound. We can also see that at an x-value of 0, the y-value is 0.

Slant Asymptotes

We previously stated that when the degree of the numerator is greater than the degree of the denominator, we will not have a horizontal asymptote. If however, the degree of the numerator is exactly one more than the degree of the denominator then there will be a slant or oblique asymptote. $$r(x) = \frac{P(x)}{Q(x)}$$ let r(x) be a rational function where the degree of the numerator is one more than the degree of the denominator. Using polynomial division, we can rewrite our function r(x) as: $$r(x) = ax + b + \frac{R(x)}{Q(x)}$$ Where the degree of R(x) is less than the degree of Q(x) and a ≠ 0. As |x| approaches infinity, the R(x)/Q(x) will approach 0. This means when |x| is large the graph of y = r(x) is going to approach the graph of the line y = ax + b. So the y = ax + b is our slant asymptote or oblique asymptote.

• Divide the numerator by the denominator and disregard the remainder
• The equation of the slant asymptote will be y = the quotient from the division

Let's look at an example.
Example #8: Find all asymptotes. $$f(x) = \frac{x^2 + x - 6}{x - 1}$$ To find the vertical asymptotes, set the denominator equal to zero and solve the resulting equation. $$x - 1 = 0$$ $$x = 1$$ The equation of the vertical asymptote is: $$x = 1$$ Since the degree of the numerator (2) is greater than the degree of the denominator (1), we know that we don't have a horizontal asymptote. However, the degree of 2 in the numerator is one more than the degree of 1 in the denominator so we will have a slant asymptote. This can be found quickly from a simple division. Recall that we will focus on the quotient only and disregard the remainder. Using the results from the synthetic division above, we can rewrite our function as: $$f(x) = \frac{x^2 + x - 6}{x - 1} = x + 2 - \frac{4}{x - 1}$$ Following our procedure, the equation of our slant asymptote is y = x + 2. We can consider where this comes from by looking at the result from our synthetic division more carefully. $$\frac{x^2 + x - 6}{x - 1} = x + 2 - \frac{4}{x - 1}$$ As |x| approaches infinity, what happens to 4/(x - 1)? It goes to 0, which means when x is very large in terms of absolute value, y approaches x + 2. Additionally, we can prove that there isn't going to be a horizontal asymptote here. Let's again divide both numerator and denominator by x², which is the highest power of x. $$f(x) = \frac{x^2 + x - 6}{x - 1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$ $$= \large{\frac{\frac{x^2}{x^2} + \frac{x}{x^2} - \frac{6}{x^2}}{\frac{x}{x^2} - \frac{1}{x^2}}}$$ $$= \large{\frac{1 + \frac{1}{x} - \frac{6}{x^2}}{\frac{1}{x} - \frac{1}{x^2}}}$$ Notice that the quotients 1/x, 6/x², and 1/x² are going to approach 0 as |x| approaches infinity. $$\frac{1 + 0 - 0}{0 - 0} \, \text{is undefined}$$ The above turns into 1/0, which is undefined, therefore, y does not approach any real number as |x| goes to infinity. This analysis shows that we will not have a horizontal asymptote in this situation. Desmos Link for More Detail

Holes in the Graph

Up to this point, we have only looked at rational expressions that are reduced to lowest terms. This means there isn't a common factor that can be canceled between numerator and denominator. What happens when this isn't the case? Let's think about this scenario with an example.
Example #9: Find the domain and sketch the graph. $$f(x) = \frac{2x^2 - 13x + 20}{x - 4}$$ First, let's consider the domain. $$x - 4 = 0$$ $$x = 4$$ $$\text{Domain:} \, \{x | x ≠ 4\}$$ Let's factor the numerator. $$f(x) = \frac{(2x - 5)(x - 4)}{(x - 4)}$$ Notice that there is a common factor of (x - 4) between the numerator and the denominator that can be canceled. $$\require{cancel}f(x) = \frac{(2x - 5)\cancel{(x - 4)}}{\cancel{(x - 4)}} = 2x - 5$$ Here is the issue, the simplified version f(x) = 2x - 5 is defined for x = 4, whereas, the original function was not. When we simplify a rational expression, we need to always carry over any domain restrictions or values for x that make the original denominator zero. So the simplified version should be written as: $$f(x) = 2x - 5, x ≠ 4$$ To graph this function, we sketch the graph of f(x) = 2x - 5 but at x = 4, we will place a hole in the graph. This will show that our function is not defined at x = 4. This hole is known as a removable discontinuity. Desmos Link for More Detail Note: if you are working on the Desmos graph, you will need to scroll your mouse over where the x-value is 4 to see the hole.

Behavior of Graphs of Rational Functions Near Vertical Asymptotes

Let f(x) be a rational function that is defined by a rational expression that is reduced to lowest terms. If n, which will be our exponent, is the largest positive integer such that: $$(x - a)^n$$ is a factor of the denominator of f(x), the graph of our rational function will look different based on whether the n-value is even or odd.

n is even

Let's look at some examples of the case where n is even. Desmos Link for More Detail Desmos Link for More Detail

n is odd

Let's look at some examples of the case where n is odd. Desmos Link for More Detail Desmos Link for More Detail

Behavior of Graphs of Rational Functions Near the Zeros of the Numerator

Recall from our lesson on graphing polynomial functions that the behavior of the graph of a polynomial function near one of its zeros will depend on whether the multiplicity of the zero is even or odd. In the case where the multiplicity of the zero is even, we saw the graph would touch the x-axis and turn around. In the case where the multiplicity of the zero is odd, we saw the graph would cross through the x-axis. The same concept will apply when graphing a rational function. Let f(x) be a rational function that is defined by a rational expression that is reduced to lowest terms. If n, which will be our exponent, is the largest positive integer such that: $$(x - k)^n$$ is a factor of the numerator of f(x), the graph of our rational function will look different based on whether the n-value is even or odd. Let's look at some examples. Desmos Link for More Detail As we can see from the graph above, at the point (1, 0), the graph touches the x-axis and then turns around. This comes from the fact that the zero of 1 has a multiplicity of 2. Desmos Link for More Detail As we can see from the graph above, at the point (5, 0), the graph crosses through the x-axis. This comes from the fact that the zero of 5 has a multiplicity of 1.

Graphing a Rational Function

Now that we have all the required information, let's look at a procedure that will allow us to make a rough sketch of a rational function. I would note that this procedure is only to be used when we can't use the techniques of function transformations with f(x) = 1/x or f(x) = 1/x² as the parent function.

Features of the Graph

• The graph should show all x and y intercepts
• The graph should show all asymptotes
• The point at which the graph intersects its nonvertical asymptote (when it exists)
• The graph should extend far enough to show the correct end behavior

The procedure given below is for a rational function that is defined by a rational expression written in lowest terms. If your rational expression is not written in lowest terms, you will need to consider the domain before simplifying. This domain will carry over to the simplified rational expression. Any x-value for which the original rational expression is undefined but is now defined in the simplified version will show up on the graph as a hole. $$f(x) = \frac{p(x)}{q(x)}$$ Where q(x) is not zero, p(x) and q(x) are polynomials, and the rational expression is written in lowest terms.

1. Find and sketch all asymptotes
2. Find the x and y intercepts
   • The y-intercept occurs at (0, f(0))
   • The x-intercepts occur at (k, 0) where k is a zero of the numerator
3. Determine whether the graph will intersect its nonvertical asymptote (if it exists)
4. Split the number line up into intervals based on the x-intercepts and vertical asymptotes
   • Pick an x-value in each interval, then find and plot the related point
5. Sketch the graph

Let's look at an example.
Example #10: Create a rough sketch of the graph. $$f(x) = \frac{x - 2}{x^2 + 2x - 3}$$ 1) Find and sketch all asymptotes. $$f(x) = \frac{x - 2}{(x - 1)(x + 3)}$$ Since the degree of the numerator (1) is less than the degree of the denominator (2), the equation of the horizontal asymptote is y = 0 or the x-axis.
The zeros of the denominator are 1 and -3, therefore the equations for the vertical asymptotes will be: $$x = 1 \, \text{and} \, x = -3$$ Note, we did not draw in the horizontal asymptote y = 0 since it is the x-axis. Feel free to draw one in on your graph if you prefer.
2) Find the x and y intercepts.
The y-intercept occurs at (0, f(0)). $$f(0) = \frac{0 - 2}{(0 - 1)(0 + 3)} = \frac{2}{3}$$ y-intercept: $$\left(0, \frac{2}{3}\right)$$ The x-intercepts occur at (k, 0) where k is a zero of the numerator. $$f(x) = \frac{x - 2}{(x - 1)(x + 3)}$$ The numerator has one zero of 2, therefore, our x-intercept will occur at (2, 0). Notice that the multiplicity of the zero is one (which is odd), so we know that the graph will cross through the x-axis at (2, 0).
x-intercept: $$(2, 0)$$ 3) Determine whether the graph will intersect its nonvertical asymptote.
The nonvertical asymptote is y = 0. So we set the function equal to zero and solve. $$\frac{x - 2}{x^2 + 2x - 3} = 0$$ $$x = 2$$ We knew this already since the x-intercept is (2, 0).
4) Split the number line up into intervals based on the x-intercepts and vertical asymptotes. Pick an x-value in each interval, then find and plot the related point. Update the graph using the points from the table. 5) Sketch the graph. Desmos Link for More Detail

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Skills Check: