Transcript Document

Section 3.2
Graphs of Equations
Objectives:
•Find the symmetries of equations with respect to x, y axis and origin.
•Use the graphical interpretation
In this presentation I also show an introduction to x-intercepts and y-intercepts of an
equation, graphically and algebraically as well as Circles
Intercepts
y-axis
d is a y-intercept ( x = 0)
Graphical Approach
d
a
b
c
x-axis
a, b, and c are x-intercepts ( y = 0)
Algebraic Approach
x-intercept: Set y = 0 and solve for x
y-intercept: Set x = 0 and solve for y
Example 1
Find the x-intercept(s) and y-intercepts(s) if they exist.
Solution:
y-axis
x-intercept(s): x = -3, 1.5, 6 and 7
1)
y-intercept(s): y = 2
2
-3
1.5
6
7
x-axis
2)
y-axis
Solution:
x-intercept(s): Does Not Exist ( D N E )
y-intercept(s): y = 3
3
-1
4
x-axis
Example 2
Find the x-intercept(s) and y-intercept(s) of the equation
x2 + y2 + 6x –2y + 9 = 0 if they exist.
Solution:
x-intercept(s): Set y = 0.
x2 + 6x + 9 = 0
( x + 3)2 = 0
x=-3
y-intercept(s): Set x = 0.
Point ( -3,0)
y2 –2y + 9 = 0
  b2  4ac  4  419  32  0
No Real Solutions
No y-intercepts
Symmetries of Graphs of Equations in x and y
Terminology
The graph is symmetric with respect
to y-axis
Graphical Interpretation
(-x,y)
The graph is symmetric with respect
to x-axis
(x,y)
(x,y)
Test for symmetry
(1) Substitution of –x for x leads to
the same equation
(2) Substitution of –y for y leads to
the same equation
(x,-y)
The graph is symmetric with respect
to origin
(x,y)
(-x,-y)
(3) Substitution of –x for x and
Substitution of –y for y
leads to the same equation
Continue…
Complete the graph of the following if
Example 3
b) Symmetric w.r.t origin
a) Symmetric w.r.t y-axis
c) Symmetry with respect to y-axis
e) Symmetric w.r.t x-axis
d) Symmetry with respect to origin
Example 4
Determine whether an equation is symmetric w.r.t y-axis, x-axis ,origin or none
a) y = 3x4 + 5x2 –4
b) y = -2x5 +4x3 +7x
Substitute x by –x and y by - y
Solution:
a)
Substitute x by –x
)4
+ 5 ( -x
3x4
5x2
y = 3( -x
=
+
)2
–4
Same equation
-4
(-y) = -2 (-x)
5
+ 4( -x )3 +7(-x)
= 2x5 – 4x3 – 7x
= - (-2x5 +4x3 +7 )
Same equation
Symmetry w.r.t
y-axis
Symmetry w.r.t origin
c) y = x3 +x2
Substitute x by –x
y = ( -x )5 + ( -x )2
= - x5 + x2
Different equation
Even if we substitute –y for
y, we get different equations
Circles
d= 2r
Equation of a circle: ( x – h )2 + ( y – k )2 = r2
Center of the circle: C( h, k )
r
r
Radius of the circle: r
Diameter of the circle: d = 2r
Example 5: Find the center and the radius of a circle whose equation is
y
( x – 3)2 + ( y + 5 )2 = 36.
Solution:
(3, 1 )
Center: C( 3, -5)
Radius: r = 6
Example 6: Graph the above Circle.
(-3, -5)
6 (3,-5)
6
6
6
(3, -11)
(9, -5)
x
Graphing Semi Circles
Upper half, Lower half, right half, and left half
Let us find the equations of the upper half, lower half, right half and left half of the circle x 2 + y2 = 25.
x2 + y2 = 25 is a circle with center ( 0, 0 ) and radius r = 5. The graph of this circle is shown
below.
To find upper and lower halves, we
solve for y in terms of x.
5
x 2  y 2  25
5
-5
y 2  25  x 2
0
-5
y   25  x 2
1) y  25  x
half plane
2
y   25  x 2  0
 0 Represents the upper
5
y
25  x 2  0
-5
5
5
-5
y   25  x 2  0
-5
Continued
…
To find right and left halves, we solve for x in terms of y.
3) x  25  y 2  0 Represents the right half
4) x   25  y 2  0 Represents the left
plane
half plane
5
x 
25  y 2  0
5
x   25  y 2  0
-5