Grade by Grade
C. Polynomial and Rational Functions · Rational Functions and Equations

Rational Equations and x-Intercepts: Solving with Graphs

C3.5About 65 minutes
Learning goals
  • I can explain the connection between the real roots of a rational equation and the x-intercepts of the corresponding rational function.
  • I can rewrite an equation f(x) = g(x) as f(x) − g(x) = 0 and connect its roots to intersection points.
  • I can explain why a zero of the numerator that is also a zero of the denominator is not a root.
  • I can use a graph to predict how many real roots an equation has, and verify roots with technology.

Words to know

rational equation
An equation that contains at least one rational expression with the variable in a denominator, for example 1/(x − 2) = 4.
root
A value of the variable that makes an equation true. The real roots of f(x) = 0 are the x-intercepts of y = f(x).
point of intersection
A point that lies on two graphs. Its x-coordinate solves the equation f(x) = g(x).
zero (of a function)
An x-value where f(x) = 0. For a rational function, a zero of the numerator that is in the domain.

Two ways to see the same answer

In the MCR3U units you solved linear and quadratic equations by connecting roots to x-intercepts: the roots of x² − 5x + 6 = 0 are the x-intercepts of y = x² − 5x + 6. The same idea works for rational equations. It gives you a picture of every solution, and it explains why some 'solutions' are fake.

Roots of f(x) = 0 are x-intercepts

For a rational function f(x) = p(x)/q(x):

  • f(x) = 0 exactly when the numerator p(x) = 0 and the denominator q(x) ≠ 0.
  • So the real roots of the equation f(x) = 0 are the x-intercepts of the graph of y = f(x).

For example, (2x − 6)/(x + 1) = 0 has the root x = 3, and the graph of y = (2x − 6)/(x + 1) crosses the x-axis at (3, 0).

Careful: a value that makes the numerator 0 but also makes the denominator 0 is not a root. (x − 2)/(x² − 4) = 0 has no roots. The numerator is 0 at x = 2, but so is the denominator, and the graph has a hole at x = 2, not an x-intercept.

Equations of the form f(x) = g(x)

To solve 1/(x − 2) = 4 with a graph, you have two choices:

  1. Graph y = 1/(x − 2) and y = 4, and find where they intersect. The x-coordinate of the intersection point is the root.
  2. Move everything to one side: 1/(x − 2) − 4 = 0. Graph y = 1/(x − 2) − 4 and find its x-intercept.

Both give x = 2.25. Combining the left side into one fraction shows why: 1/(x − 2) − 4 = (1 − 4(x − 2))/(x − 2) = (9 − 4x)/(x − 2), which is 0 when x = 9/4 = 2.25.

y = 1/(x − 2) meets y = 4xy−112345−3−2−112345x = 2y = 4(2.25, 4)y = 1/(x − 2) − 4 crosses y = 0xy−112345−7−6−5−4−3−2−11x = 2y = −4(2.25, 0)
Left: the root of 1/(x − 2) = 4 is the x-coordinate of the intersection point. Right: subtracting 4 moves the graph down 4 units, and the same root becomes an x-intercept.

Using technology

On a graphing calculator, graph the function and use the zero (or root) command to find an x-intercept, or graph both sides and use the intersect command. In a free online graphing tool, you can tap or click on an intersection or an x-intercept to see its coordinates. A computer algebra system (CAS) can solve the equation directly. In this course every answer can be found by hand, but technology is an excellent way to check your work and to predict how many roots to expect.

Worked example

Example 1: A root from the numerator

Find the real roots of (2x − 6)/(x + 1) = 0 and describe the matching feature of the graph.

  1. The fraction is 0 when the numerator is 0: 2x − 6 = 0, so x = 3.
  2. Check the denominator: 3 + 1 = 4 ≠ 0, so x = 3 is allowed.

x = 3. The graph of y = (2x − 6)/(x + 1) has an x-intercept at (3, 0).

Worked example

Example 2: When the numerator's zero is not a root

Solve (x − 2)/(x² − 4) = 0.

  1. The numerator is 0 at x = 2.
  2. But the denominator x² − 4 is also 0 at x = 2, so the expression is undefined there.
  3. Simplify: (x − 2)/((x − 2)(x + 2)) = 1/(x + 2), x ≠ 2. A fraction with numerator 1 is never 0.

No real roots. The graph of y = 1/(x + 2) with a hole at (2, 1/4) never touches the x-axis.

Worked example

Example 3: Counting roots with a graph

How many real solutions does 1/(x² + 1) = k have for k = 1.5, 1, 0.5 and −0.5?

  1. The graph of y = 1/(x² + 1) has a maximum (0, 1), is always positive, and approaches y = 0 at both ends.
  2. Draw horizontal lines y = k and count intersections.
  3. k = 1.5 is above the maximum: none. k = 1 touches the maximum: one (x = 0). k = 0.5 cuts both sides of the hill: two. k = −0.5 is below the graph: none.
  4. Check k = 0.5 algebraically: x² + 1 = 2, so x = ±1. Two roots ✓.

0, 1, 2 and 0 solutions.

How many solutions does 1/(x² + 1) = k have?xy−4−3−2−11234−0.50.511.5k = 1.5: nonek = 1: onek = 0.5: twok = −0.5: none
Sliding a horizontal line y = k up and down shows how many solutions 1/(x² + 1) = k has: none above 1, one at 1, two between 0 and 1, and none for k ≤ 0.
Worked example

Example 4: Estimating from a graph, then confirming

Use a graph to estimate the roots of x/(x − 1) = x − 3. Then confirm them algebraically.

  1. Graph y = x/(x − 1) (asymptotes x = 1 and y = 1) and the line y = x − 3. They meet twice: near x ≈ 0.7 and x ≈ 4.3.
  2. Algebra: multiply both sides by x − 1 (x ≠ 1): x = (x − 3)(x − 1) = x² − 4x + 3.
  3. So x² − 5x + 3 = 0, and the quadratic formula gives x = (5 ± √13)/2.
  4. (5 − √13)/2 ≈ 0.697 and (5 + √13)/2 ≈ 4.303, matching the estimates. Neither is 1, so both are valid.

x = (5 ± √13)/2, about 0.70 and 4.30.

y = x/(x − 1) and y = x − 3xy−2−1123456−6−5−4−3−2−112345x = 1y = 1y = x − 3x ≈ 0.70x ≈ 4.30
The line y = x − 3 crosses both branches of y = x/(x − 1), so the equation has two real roots. You will learn the algebra step (multiplying by the denominator) in detail in the next lesson.
Just for fun

I told my friend that 1/x = 0 has no solution. She suggested trying a really, really big x. I tried 1 000 000 000. The equation said, 'Close, but no.'

Hands-on activity

Slide-a-ruler solution counter

You need: graph paper, a pencil, a ruler and a calculator.

  1. Make a table for y = 1/(x² + 1) with x = −4, −3, −2, −1, −0.5, 0, 0.5, 1, 2, 3, 4. Use a large scale on the y-axis (for example 10 squares per unit).
  2. Plot the points and join them into a smooth hill.
  3. Lay the ruler horizontally across the graph at heights 1.2, 1, 0.8, 0.5, 0.2, 0 and −0.3. Each time, count where the ruler edge crosses the curve.
  4. Make a table: k and number of solutions of 1/(x² + 1) = k.
  5. For k = 0.2, read the two x-values from your graph. Then solve 1/(x² + 1) = 0.2 by hand and compare.

What to look for: y-values 0.06, 0.1, 0.2, 0.5, 0.8, 1, 0.8, 0.5, 0.2, 0.1, 0.06. Counts: 0, 1, 2, 2, 2, 0, 0. At k = 0 there are none because the graph never reaches the x-axis (it is the horizontal asymptote). For k = 0.2: x² + 1 = 5, so x = ±2.

Your turn

1

Find the real root of (3x + 12)/(x − 5) = 0.

2

Does (x − 1)/(x² − 1) = 0 have a real root? Explain.

3

Solve 2/(x + 3) = 1 by finding the x-intercept of a single rational function.

4

How many real solutions does 1/(x² − 4) = −1/8 have? Explain using the graph.

5

How many real solutions does 1/(x² − 4) = −1 have? Find them.

6

Explain how to check the root of (x + 6)/(2x − 1) = 0 with graphing technology.

7

Solve 3/(x − 1) = x + 1. How many intersection points do the two graphs have?

More practice: warm-up

1

Find the x-intercept of y = (x − 7)/(x + 2). What equation does it solve?

2

Is x = −2 a root of (x − 7)/(x + 2) = 0?

3

Find the root of (2x + 5)/(x − 1) = 0.

4

The graphs of y = f(x) and y = 3 meet at x = 4. Give a root of f(x) − 3 = 0.

5

Does 1/x = 0 have a solution?

More practice: on your own

1

Solve 5/(x + 2) = 2 and name the x-intercept that matches.

2

How many solutions does (x + 1)/(x − 1) = 1 have? Explain with the graph and with algebra.

3

Solve (x + 1)/(x − 1) = 3.

4

How many solutions does 1/(x² + 4) = k have for k = 1/4, 1/8 and 1/2?

5

Solve 1/(x − 3)² = 4. Explain why there are two roots even though the graph has only one vertical asymptote.

Challenge

1

For which values of k does 2/(x² + 1) = k have exactly two real solutions?

2

For which values of k does (x + 1)/(x − 1) = k have no solution?

3

The graphs of y = 4/x and y = 5 − x intersect twice. Find the intersection points exactly and explain the link to the x-intercepts of y = 4/x + x − 5.

Check your understanding

Quiz

1.

The real roots of a rational equation f(x) = 0 correspond to which feature of y = f(x)?

2.

What is the root of (4x − 2)/(x + 5) = 0?

3.

Why does (x + 3)/(x² − 9) = 0 have no solution?

4.

The graphs of y = f(x) and y = 2 intersect at (−1, 2) and (4, 2). What are the solutions of f(x) = 2?

5.

How many real solutions does 1/(x² + 1) = 0.5 have?

6.

Solving 1/(x − 2) = 4 is the same as finding the x-intercept of which function?

For parents and teachers

Big idea (C3.5): solving f(x) = 0 is the same as finding the x-intercepts of y = f(x), and solving f(x) = g(x) is the same as finding where the two graphs meet. A graph shows how many roots to expect and exposes 'fake' roots, such as a numerator zero that is really a hole.

How to help: If you have a phone or computer, open a free online graphing tool and type in the function your teen is working on. Ask them to predict the x-intercepts before you press enter, and to explain any missing intercepts.

Common difficulties:

  • Counting a zero of the numerator as a root even when it also zeroes the denominator.
  • Forgetting that a horizontal line y = k can meet a graph 0, 1 or 2 times (or more), so equations can have no solution.
  • Reading the y-coordinate of an intersection point as the answer instead of the x-coordinate.