Chapter 1.1 – Graphs of Equations What you should learn 1. Sketch graphs of equations 2. Find x- and y- intercepts of graphs of equations 3. Use symmetry to sketch graphs of equations 4. Find the equations of and sketch graphs of circles 5. Use graphs of equations involving real-life problems

Chapter 1.1 – Graphs of Equations Equation in two variables – a way to represent a relationship between two quantities. A solution or solution point (a,b) of an equation in x and y is true when a is substituted for x, and b is substituted for y. The graph of an equation is the set of all points that are solutions of the equation.

Example 1: Determine whether each point lies on the indicated graph : p y = x 2 – 3x + 2 a. (2,0) b. (-2, 8) 10. y = 4 – l x – 2 l a. (1,5) b. (6,0)

Solution : Determine whether each point lies on the indicated graph : p y = x 2 – 3x + 2 b. (-2, 8) a. (2,0) y = x 2 – 3x = 2 2 – 3(2) = (-2) 2 – 3(-2) = 4 – = = 0 Yes 8 = y = 4 – l x – 2 l a. (1,5) b. (6,0)

Solution 10. y = 4 – l x – 2 l a. (1,5) b. (6,0) y = 4 – l x – 2 l 5 = 4 – l 1 – 2 l 0 = 4 – l 6 – 2 l 5 = 4 – 1 0 = 4 – 4 5 = 3 no 0 = 0 Yes Not a solution (6,0) is a solution

Point-plotting method Sketching the graph of an equation by point-plotting 1. If possible, re-write the equation so that one of the variables is isolated on one side of the equation 2. Make a table of values showing several solution points 3. Plot these points on a rectangular coordinate system 4. Connect the points with a smooth curve or line

Example 2. Complete the table. Use the resulting solution points to sketch the graph of the equation. (p. 84) 16. y = ¾ x – 1 X-2014/32 Y (x,y)

Study tip Linear equation can be in the form y = mx + b, and forms a line when graphed Quadratic equation can be in the form y = ax 2 + bx + c, and forms a parabola when graphed Circle – can be in the form (x-h) 2 + (y-k) 2 = r 2 and forms a circle when graphed

Intercepts of a graph Intercepts – solution points that have zero as either x- coordinate or y-coordinate because they are the points at which the graph intersects the x- or y-axis. It is possible for a graph to have no intercepts, one intercept, or several intercepts

Intercepts No x-intercepts One y-intercept 4 x-intercepts 1 y-intercept 1 x-intercept 2 y-intercepts No intercepts

Identifying x- and y-intercepts Graphically estimate the x- and y-intercepts of the graph. (p. 84) 20. y = 16 – 4x 2 (refer to the graph on p. 84) 28. y = x 3 – 4x (refer to the graph on p. 84)

Symmetry Symmetry with respect to the x-axis means that if the Cartesian plane is folded along the x-axis, the portion of the graph above the x- axis would coincide with the portion below the x-axis. Symmetry with respect to the y-axis or the origin can be described in a similar manner. refer to (p. 80)

Algebraic tests for symmetry 1. The graph is symmetric with respect to the x-axis if replacing y with –y yields an equivalent equation 2. The graph is symmetric with respect to the y-axis if replacing x with –x yields an equivalent equation 3. The graph is symmetric with respect to the origin if replacing x with –x, and y with –y yields an equivalent equation watch video

Graphical Test for symmetry 1. A graph is symmetric with respect to the x-axis if, whenever (x,y) is on the graph, (x,y) is also on the graph 2. A graph is symmetric with respect to the y-axis if, whenever (x,y) is on the graph, (-x,y) is also on the graph 3. A graph is symmetric with respect to the origin if, whenever (x,y) is on the graph, (-x,-y) is also on the graph Watch video

Testing for symmetry Use the algebraic tests to check for symmetry with respect to both axes and the origin (p. 85) 26. x – y 2 = y = x 4 - x 2 + 3

solution 26. x – y 2 = 0 for symmetry with respect to the x-axis x – (-y) 2 = 0 x – y 2 = 0, which is equivalent to the original equation, so yes, symmetric with respect to the x- axis for symmetry with respect to the y-axis, - x – y 2 = 0, is not equivalent to the original, so not symmetric with respect to the the y-axis - x – (- y) 2 = 0 - x – y 2 = 0, is not equivalent to the original, so not symmetric with respect to the the origin

Example – sketch the graph Identify any intercepts and test for symmetry. Then sketch the graph of the equation. ( p. 85) 38. y = 2x – y = 1 - lxl

Circles - Standard form of the equation of a circle The point (x,y) lies on the circle of radius r and center (h,k) if and only if (x-h) 2 + (y-k) 2 = r 2 with center at (h,k) and radius r, and x 2 + y 2 = r 2 if the center is at the origin

Example Write the standard form of an equation of the circle with the given characteristics (p 85) 64. Center (-7, -4); radius : Center: (3,-2) ; solution point (-1, 1) 68. Endpoints of a diameter: (-4, -1), (4,1)

solution: r = √ (3 – (-1)) 2 + (-2 – 1)) 2 = √ (-3) 2 = √ 25 = 5 (x – 3) 2 + (y – (-2)) 2 = 5 2 (x – 3) 2 + (y + 2) 2 = 25 Since the radius is not given, you have to use the distance formula to find the radius of the circle

Example Find the center and radius of the circle and sketch its graph (p 85) 70. x 2 + y 2 = (x – 2) 2 + ( y + 3 ) 2 = 16 9 HOMEWORK: READ LESSON 1.4, Homework on-line is now available at