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We can draw graphs accurately by making a table of values by substituting into an algebraic equation and then plotting the points on a Cartesian plane, as above. Or we can sketch the graphs of algebraic equations by having a good knowledge of their
general shape. This may involve finding their intercepts or doing transformations of the basic curves. This sketching of curves is usually a “snapshot” of the general shape and its properties, and usually does not give as accurate a result as drawing them by a table of values. The next example uses intercepts.
Example. Sketch the graph of the parabola y = 0 — 2)0 + 3).
Solution. We know that the basic shape of the parabola is as shown in the top part of figure c. A parabola can have up to two x intercepts and one y intercept. To find the x intercepts, substitute y = 0 in the equation of the curve, and solve the resulting quadratic equation:
0 = 0- 2)0 + 3)
x = 2 and x = — 3 See Solving a Quadratic Equation
To find the y intercept, substitute x = 0 in the equation of the curve:
y = (0 - 2)(0 + 3) y = —2x3 y = -6
We fix these three points on the axes, and then sketch the general shape of the parabola through these three points.
The graph shown in figure c is a sketch of the parabola y = (x — 2)0 + 3).
This process is explained more fully in the entry Quadratic Graphs.
Conversion Graphs More information can be found in the entries Conversion and Extrapolation.
Anne’s teacher has decided to scale up her students’ mathematics marks from a recent examination, because the questions were too hard. The teacher finds the bottom, median, and top marks from the examination and in a table writes the new bottom, median, and top marks she wants her form to have. The original scores in the exam are called raw marks.
Bottom Median Top
Raw marks 20 40 70
Scaled marks 30 60 90
The points (20, 30), (40, 60), and (70, 90) are plotted on a set of axes and joined with two straight-line segments. These lines constitute the conversion graph (see figure d). Anne had a raw mark of 65%, and her scaled mark can be found by locating 65% on the raw marks axis and drawing a vertical line up to the line of the graph. From this point on the graph draw a horizontal line onto the new marks axis, and read off the value 85%. Anne’s scaled mark is 85%.
Circle Graphs More information can be found in the entry Conic Sections. If a circle is drawn that has its center at the origin (0, 0) and with a radius of r, the equation of its graph is x2 + y2 = r2. The process of drawing the graph of a circle is demonstrated by the following example.
Example. Draw a circle whose equation is x2 + y2 = 9.
Solution. Compare the equations2 + y2 = 9 with the formula x2 + y2 = r2, which is a circle whose center is at the origin (0, 0) and radius r. Write
r2 = 9
r = V9 Taking the square root
r — 3
We now draw the circle of radius 3 and center at (0, 0).
Using Graphs to Solve Algebraic Equations Equations can be solved using algebra; see the entries Linear Equation and Quadratic Equations. We now leam how to solve equations using graphs, but this method is not expected to give an accurate solution to the equation.
The theory for this graphical method is that if we draw two graphs y = fix) and y = g(x) on the same axes and from the graphs read off the x values where the two graphs intersect, then those values of x are the solutions of the equation fix) = g(x).
Example. Use a graphical method to solve the equation 2X = x2. This equation can also be written as 2X — x2 = 0.
Solution. The two graphs we need to draw are y = 2X and y = x2 on the same axes, and the more accurately the graphs are drawn, the more accurate will be the solutions to the equation. We will use a table of values for each graph. First we have
y = 2x (2 dp) 0.13 0.25 0.5 1 2 4 8
Plotting the coordinates (—3, 0.13), (—2, 0.25), (—1, 0.5), (0, 1), (1, 2), (2, 4), and (3, 8) on the axes gives the graph of y = 2X:
y=x2 9 4 1 0 1 49
Plotting the coordinates (—3, 9), (—2, 4), (—1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) on the axes gives the graph of y = x2.
The final stage of solving the equation 2X = x2 is to read off from the graph, as accurately as one can, the values of x where the graphs intersect (figure f). There are two values, and the approximate solution of the equation is x = 2 and x = —0.8.
References: Asymptote, Cartesian Coordinates.
The symbol for “greater than” is >. The number 5 is greater than the number —3, since 5 is to the right of —3 on the number line. This is written as 5 > —3.
The symbol for “less than” is <. The number —3 is less than the number 5, since —3 is to the left of 5 on the number line. This is written as —3 < 5.
Example 1. Find a value for x where x < —2.
Solution. We can choose any number to the left of —2 on the number line, say —6 or —2.1. All numbers represented by the continuous line in figure a are solutions. Note there is a “hole” in this line at the point —2 itself, because —2 is not included in the set of solutions.