 # Precalculus Part I: Orientation to Functions Day 1: The Cartesian Plane Day 2: Graphing Equations.

## Presentation on theme: "Precalculus Part I: Orientation to Functions Day 1: The Cartesian Plane Day 2: Graphing Equations."— Presentation transcript:

Precalculus Part I: Orientation to Functions Day 1: The Cartesian Plane Day 2: Graphing Equations

Key Skill: Plot a given equation. Preliminary Consideration: Is the equation suitable for the Cartesian plane? Yes, if the number of explicit variables displayed in the equation is either 1 or 2.

Examples Not suitable for the Cartesian plane: 2 + 3 = 5  no explicit variables x 2 + y 2 + z 2 = 27  3 explicit variables: x, y, z Suitable for the Cartesian plane: x = 3y – 2x = 5 y = -2x 2 + y 2 = 25

Example Consider the equation: y – 2x = 5 Preliminary: Is the equation suitable for the Cartesian plane? The explicit variables are y and x. There are two explicit variables. Hence, the equation is suitable for the Cartesian plane.

Graphing Equations There are two methods of plotting a given equation: Point Plotting and Sketching by key characteristics of the equation

Given an equation: y – 2x = 5 and a point: (1,2) Does the point fit the equation? Answer: 2 – 2(1) = 5? So, do not plot the point (1,2). Point Plotting Yes No

Given an equation: y – 2x = 5 and a point: (-2,1) Does the point fit the equation? Answer: 1 – 2(-2) = 5? So, plot the point (-2,1) Continue until you have plotted enough points to get a grip on the graph of the equation. Point Plotting Yes No

Organized point-plotting Step 1: Make a table of friendly values. x -2 0 1 2 y 1 3 5 7 9 Step 2: Plot the points. x y Step 3: Smoothly join the points. x y y = 2x+5 Graph y = 2x + 5

Key Characteristics Intercepts Vertical intercept (Set x = 0) Horizontal intercept (Set y = 0) Symmetry About the vertical axis (Replace x by –x) About the horizontal axis (Replace y by –y) About the origin (Replace x, y, by –x, -y)

Intercepts Determine the intercepts, those points where the graph crosses the axes. y-intercept: graph crosses the y-axis. Set x = 0 and solve for y: y = 2x + 5 y = 2(0) + 5 y = 5 Hence, the y-intercept is (0,5). x y (0,?)

Intercepts x-intercept: graph crosses the x-axis. Set y = 0 and solve for x: y = 2x + 5 0 = 2x + 5 x = -5/2 Hence, the x-intercept is (-5/2,0). x y (?,0)

Symmetry Test y = x 3 – 2x for symmetry 1. About the vertical axis Replace x by –x. If the equation is equivalent to the original, then there is symmetry about the vertical axis. y = (-x) 3 – 2(-x) y = -x 3 + 2x This is not equivalent to the original, so no symmetry about the vertical axis.

Symmetry Test y = x 3 – 2x for symmetry 2. About the horizontal axis Replace y by –y. If the equation is equivalent to the original, then there is symmetry about the horizontal axis. -y = x 3 – 2x y = -x 3 + 2x This is not equivalent to the original, so no symmetry about the horizontal axis.

Symmetry Test y = x 3 – 2x for symmetry 3. About the origin Replace x, y by –x, and –y, respectively. If the equation is equivalent to the original, then there is symmetry about the origin. -y = (-x) 3 – 2(-x) -y = -x 3 + 2x y = x 3 – 2x This is equivalent to the original, so the graph of the equation is symmetric about the origin.

WinPlot

Another Example: x 2 (y+1) = 1

Download ppt "Precalculus Part I: Orientation to Functions Day 1: The Cartesian Plane Day 2: Graphing Equations."

Similar presentations