Post on 11-Aug-2020
Graphing Equations
Chapter 2: Section 2
Math 1111: College Algebra
September 19, 2011
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Equations for Circles
Question: How do we find the equation for a circle?
-10 -5 5
-8
-6
-4
-2
2
4
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
A Circle’s Relationship with its Radius
Suppose we begin with the point (h, k) and wish to find all of theelements that are a distance of r away from the aforementionedpoint.
Graphically this leads to us forming a circle centered at thepoint (h, k) and a radius of r . Since r represents a distance, wecan use the distance formula to find an equation of all the points(x , y) that are of distance r away from the point (h, k). Hence theequation for a circle can be written as follows:√
(x − h)2 + (y − k)2 = r .
More simply put, the equation for a circle is given by
(x − h)2 + (y − k)2 = r2,
where (h, k) is the center of the circle and r is the radius.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
A Circle’s Relationship with its Radius
Suppose we begin with the point (h, k) and wish to find all of theelements that are a distance of r away from the aforementionedpoint. Graphically this leads to us forming a circle centered at thepoint (h, k) and a radius of r .
Since r represents a distance, wecan use the distance formula to find an equation of all the points(x , y) that are of distance r away from the point (h, k). Hence theequation for a circle can be written as follows:√
(x − h)2 + (y − k)2 = r .
More simply put, the equation for a circle is given by
(x − h)2 + (y − k)2 = r2,
where (h, k) is the center of the circle and r is the radius.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
A Circle’s Relationship with its Radius
Suppose we begin with the point (h, k) and wish to find all of theelements that are a distance of r away from the aforementionedpoint. Graphically this leads to us forming a circle centered at thepoint (h, k) and a radius of r . Since r represents a distance, wecan use the distance formula to find an equation of all the points(x , y) that are of distance r away from the point (h, k).
Hence theequation for a circle can be written as follows:√
(x − h)2 + (y − k)2 = r .
More simply put, the equation for a circle is given by
(x − h)2 + (y − k)2 = r2,
where (h, k) is the center of the circle and r is the radius.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
A Circle’s Relationship with its Radius
Suppose we begin with the point (h, k) and wish to find all of theelements that are a distance of r away from the aforementionedpoint. Graphically this leads to us forming a circle centered at thepoint (h, k) and a radius of r . Since r represents a distance, wecan use the distance formula to find an equation of all the points(x , y) that are of distance r away from the point (h, k). Hence theequation for a circle can be written as follows:√
(x − h)2 + (y − k)2 = r .
More simply put, the equation for a circle is given by
(x − h)2 + (y − k)2 = r2,
where (h, k) is the center of the circle and r is the radius.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
A Circle’s Relationship with its Radius
Suppose we begin with the point (h, k) and wish to find all of theelements that are a distance of r away from the aforementionedpoint. Graphically this leads to us forming a circle centered at thepoint (h, k) and a radius of r . Since r represents a distance, wecan use the distance formula to find an equation of all the points(x , y) that are of distance r away from the point (h, k). Hence theequation for a circle can be written as follows:√
(x − h)2 + (y − k)2 = r .
More simply put, the equation for a circle is given by
(x − h)2 + (y − k)2 = r2,
where (h, k) is the center of the circle and r is the radius.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3). We also find thepoint (5,−3) lies on the circle. As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6. This gives us
h = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3).
We also find thepoint (5,−3) lies on the circle. As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6. This gives us
h = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3). We also find thepoint (5,−3) lies on the circle.
As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6. This gives us
h = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3). We also find thepoint (5,−3) lies on the circle. As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6.
This gives ush = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3). We also find thepoint (5,−3) lies on the circle. As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6. This gives us
h = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining the Equation of a Circle
Question: What is the equation for the circle pictured below?
-10 -5 5
-8
-6
-4
-2
2
4
Notice that the center is located at (−1,−3). We also find thepoint (5,−3) lies on the circle. As a result we find the radius isgiven by r =
√(−1− 5)2 + (−3− (−3))2 = 6. This gives us
h = −1, k = −3, and r = 6, so we can use our equation for a circleto find the equation is given by the following equations:
(x − (−1))2 + (y − (−3))2 = 62
(x + 1)2 + (y + 3)2 = 36
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Determining Circles from Equations
Question: Is the following equation a circle?
x2 + y2 − 6x + 10y + 13 = 0
Answer: We begin by grouping all of the terms involving x , y , andconstants together as follows:
(x2 − 6x) + (y2 + 10y) = −13.
We now try to create perfect squares with the x and y terms giving
(x2−6x+
(−6
2
)2
)+(y2+10y+
(10
2
)2
) = −13+
(−6
2
)2
+
(10
2
)2
.
Since we have perfect squares in place we can rewrite the equationas
(x − 3)2 + (y + 5)2 = 21.
This is a circle centered at (3,−5) with radius√
21.Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Types of Symmetry
The graph of an equation is symmetric to the x-axis if forevery ordered pair (a, b) on the graph, we find (a,−b) is alsoon the graph.
The graph of an equation is symmetric to the y-axis if forevery ordered pair (a, b) on the graph, we find (−a, b) is alsoon the graph.
The graph of an equation is symmetric to the origin if forevery ordered pair (a, b) on the graph, we find (−a,−b) isalso on the graph.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Types of Symmetry
The graph of an equation is symmetric to the x-axis if forevery ordered pair (a, b) on the graph, we find (a,−b) is alsoon the graph.
The graph of an equation is symmetric to the y-axis if forevery ordered pair (a, b) on the graph, we find (−a, b) is alsoon the graph.
The graph of an equation is symmetric to the origin if forevery ordered pair (a, b) on the graph, we find (−a,−b) isalso on the graph.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Types of Symmetry
The graph of an equation is symmetric to the x-axis if forevery ordered pair (a, b) on the graph, we find (a,−b) is alsoon the graph.
The graph of an equation is symmetric to the y-axis if forevery ordered pair (a, b) on the graph, we find (−a, b) is alsoon the graph.
The graph of an equation is symmetric to the origin if forevery ordered pair (a, b) on the graph, we find (−a,−b) isalso on the graph.
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1.
Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation.
So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1.
This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1.
Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1
=⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the x-axis
Consider the equation given by x = y2− 1. Let us try to determinethe types of symmetry for this equation. So we begin by assumingthat (a, b) is a point on the graph of x = y2 − 1. This impliesa = b2 − 1. Now we apply the points listed in the three scenariosfrom the previous slides to see if they satisfy the aforementionedequation. For the x-axis we find
a = (−b)2 − 1 =⇒ a = b2 − 1
After trying the other types of symmetries, the only symmetry tosimplify back to the original is symmetry about the x-axis. Thegraph of the equation verifies our conclusion
-2 2 4
-2
-1
1
2
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
.
Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis.
So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x.
This implies b = aa3+2a
. For symmetry to the y-axis weplug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a.
For symmetry to the y-axis weplug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)
=⇒ b =−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)
=⇒ b =a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the y-axis
Consider the equation given by y = xx3+2x
. Let us try to determineif the graph of this equation is symmetric to the y-axis. So webegin by assuming that (a, b) is a point on the graph ofy = x
x3+2x. This implies b = a
a3+2a. For symmetry to the y-axis we
plug in −a for x and b for y to obtain
b =−a
(−a)3 + 2(−a)=⇒ b =
−a
−a3 − 2a
b =−a
−(a3 + 2a)=⇒ b =
a
a3 + 2a
This implies the graph is symmetric to the y-axis as we can see inthe following graph.
-4 -2 2 4
-0.2
0.2
0.4
0.6
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a.
Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a
−b = −(a3 + 2a) =⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a. Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a)
=⇒ −b = −a3 − 2a
−b = −(a3 + 2a) =⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a. Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a
−b = −(a3 + 2a) =⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a. Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a
−b = −(a3 + 2a)
=⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a. Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a
−b = −(a3 + 2a) =⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1
Graphing EquationsEquations for CirclesSymmetry
Symmetric to the origin
Consider the equation y = x3 + 2x . In order to check if this graphis symmetric to the origin we assume (a, b) is on the graph of theaforementioned equation which gives b = a3 + 2a. Now plug in −afor x and −b for y to find
−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a
−b = −(a3 + 2a) =⇒ b = a3 + 2a
Hence the equation y = x3 + 2x is symmetric to the origin withthe graph given below.
-3 -2 -1 1 2 3
-10
-5
5
10
Moore Chapter 1