Lecture 05
Instructor: Asgher Ali
Feb 01, 2023
1 Graph
1.1 Sketching a Graph by Plotting Points
We can sketch the graph of any linear equation by considering some of
the points and join them with the help of a straight line.
Example 1.1.1 Sketch the graph of the equation 2x − y = 3
1. We first transform the equation in terms of y on the LHS.
2. We design a table.
x y = 2x − 3 (x, y)
-3 -9 (-3,-9)
-2 -7 (-2,-7)
-1 -5 (-1,-5)
0 -3 (0,-3)
1 -1 (1,-1)
2 1 (2,1)
3 3 (3,3)
3. Plot the points on the graph
1
Example 1.1.2 Sketch the graph of the equation y = x2 − 2
1. We first transform the equation in terms of y on the LHS.
2. We design a table.
x y = x2 − 2 (x, y)
-3 7 (-3,7)
-2 2 (-2,2)
-1 -1 (-1,-1)
0 -2 (0,-2)
1 -1 (1,-1)
2 2 (2,2)
3 7 (3,7)
3. Plot the points on the graph
2
1.2 Graphing the Absolute Value Equation
Example 1.2.1 Sketch the graph of the equation y = |x|.
1. We first transform the equation in terms of y on the LHS.
2. We design a table.
x y = |x| (x, y)
-3 3 (-3,3)
-2 2 (-2,2)
-1 1 (-1,1)
0 0 (0,0)
1 1 (1,1)
2 2 (2,2)
3 3 (3,3)
3. Plot the points on the graph
3
1.3 Intercepts
The x-coordinates of the points where a graph intersects the x-axis are
called the x − intercepts of the graph and are obtained by setting
y = 0 in the equation of the graph.
Example 1.3.1 Find the x-intercepts of the graph of the
equation y = x2 − 2
1. To find the x intercepts,√we set y = 0 and solve for
√ x. In
√this
example, we get x = ± 2. The x intercepts are 2, − 2.
2. To find the y intercepts, we set x = 0 and solve for y. In this
example, we get y = −2.
2 Circles
How to find an equation of a graph? If a geometric curve can
be represented by an algebraic equation, the rules of algebra can be
used to analyze the curve.
4
As an example, lets find the equation of a circle with radius r and
center (h, k). Circle is the set of all points P (x, y) whose
distance from the center C(h, k) is a constant r. Thus
P is on the circle if and only if d(P, C) = r. From the distance
fromula, we have
r
(x − h)2 + (y − k)2 = r
(x − h)2 + (y − k)2 = r2
2.1 Graphing a Circle
Graph each equation.
1. x2 + y 2 = 25
2. (x − 2)2 + (y + 1)2 = 25
2.2 Finding an Equation of Circle
1. Find an equation of the circle with radius 3 and center (2, −5).
2. Find an equation of the circle that has the points P (1, 8) and
Q(5, −6) as the end point of a diameter.
The answers are
1. (x − 2)2 + (y + 5)2 = 9
2. (x − 3)2 + (y − 1)2 = 53
2.3 Identifying an Equation of a Circle
Show that the equation x2 + y 2 + 2x − 6y + 7 = 0 represents a
circle, and find the center and radius of the circle.
5
3 Symmetry
Sometimes the part of the graph to the left of the y−axis is the mirror
image of the part to the right of the y−axis. The reason is that if
the point (x, y) is on the graph, so is (−x, y), and these points are
reflections of each other about the y−axis. In this situation, we say
that the graph is symmetric with respect to the y-axis.
Test the equation for symmetry.
1. y = x4 + x2
2. x = y 4 − y 2
3. y = x3 + 10x
