Graph Both Functions to Find Solutions to the Equation
Solving Equations Graphically |
Contents: This page corresponds to § 2.2 (p. 187) of the text.
Suggested Problems from Text:
p. 187 #2, 4, 8, 9, 15, 16, 28, 29, 31, 32, 34, 35, 39, 41, 45, 51, 53, 56, 63, 69
Zeros of Functions
Solving Equations Graphically
Intersection Points of Graphs
Zeros of Functions
A zero of a function f is a number a such that f(a) = 0. In other words, a zero of f is a solution for the equation f(x) = 0. Furthermore, x = a is a zero of f if and only if the point (a,0) is an x-intercept of the graph of f.
Given any equation in x, there is an equivalent equation of the form f(x) = 0. Solving the equation is the same as finding the zeros of f, which is the same as finding the x-intercepts of the graph of the equation.
Example 1.
Problem: Solve 5x + 2 = 3x - 7
Subtracting 3x and adding 7 to both sides of the equation yields the equivalent equation:
2x + 9 = 0.
So, the following three problems are equivalent:
Solve 5x + 2 = 3x - 7.
Find the zeros of f(x) = 2x + 9.
Find the x-intercepts of y = 2x + 9.
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Solving Equations Graphically
The fact that solving equations may be thought of as finding the x-intercepts of a graph makes graphing utilities very useful for equation solving. You should understand that in most cases, a graphing utility will not find the exact solutions of an equation, merely approximations.
Example 2.
Approximate the solutions for x3 + 5x = 2x2 + 7.
The first step is to rewrite the equation in "f(x) = 0 form"; iI.e., move all the terms to one side of the equation. Using the notation required by the Java Grapher, we have
x^3 + 5*x - 2*x^2 - 7 = 0.
Now graph the function f(x) = x^3 + 5*x - 2*x^2 - 7.
There are several things to notice about the graph of f.
1. The y-intercept appears to be (0,-7), and in fact these are the exact coordinates. The y-intercept of the graph of a function is easy to find. You simply evaluate the function at 0, and in this example, f(0) = -7. Of course, if 0 is not in the domain of the function, then there is no y-intercept.
2. In this example there appears to be only one x-intercept with first coordinate approximately 1.5. There could be others that are outside of the viewing rectangle, but in this case there are not.
3. If you use the Table feature to evaluate the function f at 1.5 you find f(1.5) = -0.625, which is not very close to 0, so 1.5 is not a very good approximation.
Use the Trace feature to get a better idea of where the x-intercept is. As the trace point gets close to the intercept, watch the y coordinate window. Remember that the goal is to find a trace point (point on the graph) with y coordinate 0. Usually you do not hit 0 exactly, so you want to note where the y coordinates change sign. In this example they go from negative to positive because the graph is increasing. The table below shows the two trace points closest to x-axis.
x
y
1.6
-0.024
1.6666667
0.4074074
This tells us that the solution for the equation is between x = 1.6 and x = 1.6666667. At this point we know that the first decimal place is a 6 but we do not know the second.
Notice that this graph is quite "steep." Because of this steepness, simply zooming in will not improve the accuracy of our approximation very quickly. Using a Zoom Box will give better results. The idea is to drag the box so that the graph is not as steep. A "tall and thin" box like the one pictured below is what we want. If the lower left-hand corner and upper right-hand corner of the box touch the graph, then when the viewing rectangle is changed, the visible portion of the graph will look like a diagonal in the viewing widow, although not perfectly straight.
The advantage of having a graph that is not as steep is that when we trace, the change in x does not result in as large a change in y. After adjusting the viewing rectangle with the Zoom Box pictured above, the two trace points closest to the x-axis are listed in the table below.
x
y
1.6
-0.024
1.6044445
0.0039665075
We now know that the solution is between x = 1.60 and x = 1.6044445. Therefore, the second decimal place is 0. Furthermore, if we round to two decimal places, then the result is 1.60 because the third decimal place is at most 4, so we would not round up.
So, our approximation of a solution for the equation is x = 1.60, and this is the value of the exact answer rounded to two decimal places.
Exercise 1:
Find the three solutions for x3 + 1 = 3x, accurate when rounded to two decimal places. Answer
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Intersection Points of Graphs
Let f and g be functions and suppose that the graphs of f and g intersect at the point (a,b). Since (a,b) is on the graph of f, b = f(a). (All points on the graph of f are of the form (x,f(x)).) But (a,b) is also on the graph of g, so b = g(a). Therefore, f(a) = g(a); i.e., a is a solution for the equation f(x) = g(x).
Example 3.
Let f(x) = x2 + 1, and g(x) = x + 3. The graphs intersect in two points, and the first coordinates of the intersection points are -1 and 2.
The equation x2 + 1 = x + 3 has two solutions, x = -1 and x = 2.
Once you know the x coordinates of the intersection points, then you find the y coordinates by evaluating either of the functions at the x values.
In this example, g(-1) = f(-1) = 2 and g(2) = f(2) = 5, so the intersection points are (-1,2) and (2,5).
It is possible to use the "Zoom and Trace" method with a graphing utility to approximate intersection points of the graphs of functions f and g. In general, it is better to find the x intercepts of the graph of h(x) = f(x) - g(x).
Example 4.
Find the points of intersection of the graphs of f(x) = x2 + 1 and g(x) = 3 - x/2.
Let h(x) = f(x) - g(x) = (x2 + 1)-(3 - x/2) = x2 - 2 + x/2. The graphs of f, g, and h are pictured below.
The two intersection points are circled. Notice that the x-intercepts of the graph of h correspond to the first coordinates of the two intersection points.
The first coordinates of the x-intercepts of h are approximately -1.69 and 1.19.
Note that these are not exact values. If we evaluate f and g at these numbers, we will get slightly different values. Since g is the simpler function, we will use g.
g(-1.69) = 3.845 and g(1.19) = 2.405. So the intersection points are approximately
(-1.69, 3.85) and (1.19, 2.41).
Exercise 2:
Approximate the points of intersection of the graphs of f(x) = x2 - 2x and g(x) = x + 1. Answer
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Graph Both Functions to Find Solutions to the Equation
Source: http://dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/IZS/gsolve/gsolve.html