College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson

Equations and Inequalities 1

Solving Equations Graphically 1.4

Solving Equations Graphically In this section, we use graphs to solve equations. To do this, we must first draw a graph using a graphing device. So, we begin by giving a few guidelines to help us use graphing devices effectively.

Solving Equations Algebraically In Chapter 1, we learned how to solve equations. To solve an equation like 0 = 3x – 5, we used the algebraic method. This means we used the rules of algebra to isolate x on one side of the equation.

Solving Equations Algebraically We view x as an unknown and we use the rules of algebra to isolate it on one side of the equation. For this equation we add 5 and then divide by 3 to obtain the solution x = 5/3.

Solving Equations Graphically We can also solve this equation by the graphical method. We view x as a variable and sketch the graph of the equation y = 3x – 5. Different values for x give different values for y. Our goal is to find the value of x for which y = 0.

Solving Equations Graphically From the graph, we see that y = 0 when x ≈ 1.7. Thus, the solution is x ≈ 1.7. Note that, from the graph, we obtain an approximate solution.

Solving Equations Graphically We summarize these methods here.

Algebraic Method—Advantages The advantages of the algebraic method are that: It gives exact answers. The process of unraveling the equation to arrive at the answer helps us understand the algebraic structure of the equation.

Algebraic Method—Disadvantage On the other hand, for many equations, it is difficult or impossible to isolate x.

Graphical Method—Advantages The graphical method gives a numerical approximation to the answer. This is an advantage when a numerical answer is desired. For example, an engineer might find an answer expressed as x ≈ 2.6 more immediately useful than x =.

Graphical Method—Advantages Also, graphing an equation helps us visualize how the solution is related to other values of the variable.

E.g. 1—Solving an Equation Algebraically and Graphically Solve the quadratic equations algebraically and graphically. x 2 – 7 = 0

E.g. 1—Solving Algebraically There is no real solution. (Add 7)

E.g. 1—Solving Graphically We graph the equations y = x 2 – 7 By determining the x-intercepts of the graphs, we find the following solutions.

E.g. 1—Solving Graphically We see that the graph crosses the x ≈ 2.6 and x ≈ –2.6

E.g. 2—Another Graphical Method Solve the equation algebraically and graphically: 5 – 3x = 8x – 20

E.g. 2—Algebraic Solution 5 – 3x = 8x – 20 – 3x = 8x – 25 (Subtract 5) –11x = –25 (Subtract 8x)

E.g. 2—Graphical Solution We could: Move all terms to one side of the equal sign. Set the result equal to y. Graph the resulting equation.

E.g. 2—Graphical Solution However, to avoid all that algebra, we graph two equations instead: y 1 = 5 – 3x and y 2 = 8x – 20 The solution of the original equation will be the value of x that makes y 1 equal to y 2. That is, the solution is the x-coordinate of the intersection point of the two graphs.

E.g. 2—Graphical Solution Using the TRACE feature or the intersect command on a graphing calculator, we see that the solution is x ≈ 2.27.

Solving Equations Graphically In the next example, we use the graphical method to solve an equation that is extremely difficult to solve algebraically.

E.g. 3—Solving an Equation in an Interval Solve the equation in the interval [1, 6]. We need to find all solutions x that satisfy 1 ≤ x ≤ 6. So, we will graph the equation in a viewing rectangle for which the x-values are restricted to this interval.

E.g. 3—Solving an Equation in an Interval The figure shows the graph of the equation in the viewing rectangle [1, 6] by [–5, 5]. There are two x-intercepts in this rectangle.

E.g. 3—Solving an Equation in an Interval Zooming in, we see that the solutions are: x ≈ 2.18 and x ≈ 3.72

Solving an Equation in an Interval The equation in Example 3 actually has four solutions. You are asked to find the other two in Exercise 32.

E.g. 4—Intensity of Light Two light sources are 10 m apart. One is three times as intense as the other. The light intensity L (in lux) at a point x meters from the weaker source is given by:

E.g. 4—Intensity of Light Find the points at which the light intensity is 4 lux. We need to solve the equation

E.g. 4—Intensity of Light The graphs of are shown. Zooming in (or using the intersect command), we find two solutions: x ≈ and x ≈

E.g. 4—Intensity of Light So, the light intensity is 4 lux at points that are 1.67 m and 7.19 m from the weaker source.