Chapter 2: Section 2 - Georgia Southern University · 2011-09-20 · Chapter 2: Section 2 Math 1111: College Algebra September 19, 2011 Moore Chapter 1. Graphing Equations Equations

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?

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-8

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-2

2

4

Moore Chapter 1


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



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 .


(x − h)2 + (y − k)2 = r2,


Moore Chapter 1



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 .


(x − h)2 + (y − k)2 = r2,


Moore Chapter 1



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 .


(x − h)2 + (y − k)2 = r2,


Moore Chapter 1



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 .


(x − h)2 + (y − k)2 = r2,


Moore Chapter 1


Determining the Equation of a Circle

Question: What is the equation for the circle pictured below?

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-8

-6

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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




-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


(x − (−1))2 + (y − (−3))2 = 62

(x + 1)2 + (y + 3)2 = 36

Moore Chapter 1




-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


(x − (−1))2 + (y − (−3))2 = 62

(x + 1)2 + (y + 3)2 = 36

Moore Chapter 1




-10 -5 5

-8

-6

-4

-2

2

4


√(−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




-10 -5 5

-8

-6

-4

-2

2

4


√(−1− 5)2 + (−3− (−3))2 = 6. This gives us


(x − (−1))2 + (y − (−3))2 = 62

(x + 1)2 + (y + 3)2 = 36

Moore Chapter 1




-10 -5 5

-8

-6

-4

-2

2

4


√(−1− 5)2 + (−3− (−3))2 = 6. This gives us


(x − (−1))2 + (y − (−3))2 = 62

(x + 1)2 + (y + 3)2 = 36

Moore Chapter 1


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




x2 + y2 − 6x + 10y + 13 = 0


(x2 − 6x) + (y2 + 10y) = −13.


(x2−6x+

(−6

2

)2

)+(y2+10y+

(10

2

)2

) = −13+

(−6

2

)2

+

(10

2

)2

.


(x − 3)2 + (y + 5)2 = 21.


21.

Moore Chapter 1




x2 + y2 − 6x + 10y + 13 = 0


(x2 − 6x) + (y2 + 10y) = −13.


(x2−6x+

(−6

2

)2

)+(y2+10y+

(10

2

)2

) = −13+

(−6

2

)2

+

(10

2

)2

.


(x − 3)2 + (y + 5)2 = 21.


21.

Moore Chapter 1




x2 + y2 − 6x + 10y + 13 = 0


(x2 − 6x) + (y2 + 10y) = −13.


(x2−6x+

(−6

2

)2

)+(y2+10y+

(10

2

)2

) = −13+

(−6

2

)2

+

(10

2

)2

.


(x − 3)2 + (y + 5)2 = 21.


21.

Moore Chapter 1




x2 + y2 − 6x + 10y + 13 = 0


(x2 − 6x) + (y2 + 10y) = −13.


(x2−6x+

(−6

2

)2

)+(y2+10y+

(10

2

)2

) = −13+

(−6

2

)2

+

(10

2

)2

.


(x − 3)2 + (y + 5)2 = 21.


21.

Moore Chapter 1




x2 + y2 − 6x + 10y + 13 = 0


(x2 − 6x) + (y2 + 10y) = −13.


(x2−6x+

(−6

2

)2

)+(y2+10y+

(10

2

)2

) = −13+

(−6

2

)2

+

(10

2

)2

.


(x − 3)2 + (y + 5)2 = 21.


21.Moore Chapter 1


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


Types of Symmetry




Moore Chapter 1


Types of Symmetry




Moore Chapter 1


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

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1

2

Moore Chapter 1



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


-2 2 4

-2

-1

1

2

Moore Chapter 1



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


-2 2 4

-2

-1

1

2

Moore Chapter 1



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


-2 2 4

-2

-1

1

2

Moore Chapter 1



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


-2 2 4

-2

-1

1

2

Moore Chapter 1




a = (−b)2 − 1 =⇒ a = b2 − 1


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-2

-1

1

2

Moore Chapter 1




a = (−b)2 − 1 =⇒ a = b2 − 1


-2 2 4

-2

-1

1

2

Moore Chapter 1


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.

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-0.2

0.2

0.4

0.6

Moore Chapter 1




. 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




b =−a

(−a)3 + 2(−a)=⇒ b =

−a

−a3 − 2a

b =−a

−(a3 + 2a)=⇒ b =

a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1




. 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


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-0.2

0.2

0.4

0.6

Moore Chapter 1






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


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1








b =−a

(−a)3 + 2(−a)

=⇒ b =−a

−a3 − 2a

b =−a

−(a3 + 2a)=⇒ b =

a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1








b =−a

(−a)3 + 2(−a)=⇒ b =

−a

−a3 − 2a

b =−a

−(a3 + 2a)=⇒ b =

a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1








b =−a

(−a)3 + 2(−a)=⇒ b =

−a

−a3 − 2a

b =−a

−(a3 + 2a)

=⇒ b =a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1








b =−a

(−a)3 + 2(−a)=⇒ b =

−a

−a3 − 2a

b =−a

−(a3 + 2a)=⇒ b =

a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1








b =−a

(−a)3 + 2(−a)=⇒ b =

−a

−a3 − 2a

b =−a

−(a3 + 2a)=⇒ b =

a

a3 + 2a


-4 -2 2 4

-0.2

0.2

0.4

0.6

Moore Chapter 1


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



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


-3 -2 -1 1 2 3

-10

-5

5

10

Moore Chapter 1




−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a

−b = −(a3 + 2a) =⇒ b = a3 + 2a


-3 -2 -1 1 2 3

-10

-5

5

10

Moore Chapter 1




−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a

−b = −(a3 + 2a)

=⇒ b = a3 + 2a


-3 -2 -1 1 2 3

-10

-5

5

10

Moore Chapter 1




−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a

−b = −(a3 + 2a) =⇒ b = a3 + 2a


-3 -2 -1 1 2 3

-10

-5

5

10

Moore Chapter 1




−b = (−a)3 + 2(−a) =⇒ −b = −a3 − 2a

−b = −(a3 + 2a) =⇒ b = a3 + 2a


-3 -2 -1 1 2 3

-10

-5

5

10

Moore Chapter 1

Documents

Chapter 2: Section 2 - Georgia Southern University · 2011-09-20 · Chapter 2: Section 2 Math 1111: College Algebra September 19, 2011 Moore Chapter 1. Graphing Equations Equations