Category Archives: Algebra

Find the center and the radius from the equation

A circle is the collection of points that are equidistant to a center point.  The distance is the radius denoted r.  The center is denoted (h, k).

The standard form of an equation of a circle is (x-h)^2+(y-k)^2=r^2 where r is the radius and   (h, k) is the center.

 

Example:  Find the center and the radius from the given equation.

(x-2)^2+(y-3)^2=16

Solution:  If you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=2.  In the binomial with y, the number after the minus sign is k.  Thus, k=3.  The r^2 lines up with 16.  So r^2=16 which means r=4.

This is the equation of a circle with a center of (2, 3) and a radius of r=4 .

 

Example:  Find the center and the radius from the given equation.

(x+1)^2+(y-4)^2=25

Solution:  Notice that one of the binomials has a + instead of the – that is in the standard form of the circle.  Rewrite the addition as subtraction.

x+1 is the same as x-(-1).

The rewritten equation is as follows.

(x-(-1))^2+(y-4)^2=25

Then, if you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=-1.  In the binomial with y, the number after the minus sign is k.  Thus, k=4.  The r^2 lines up with 25.  So r^2=25 which means r=5.

This is the equation of a circle with a center of (-1, 4) and a radius of r=5 .

 

Example:  Find the center and the radius from the given equation.

x^2+(y+7)^2=12

Solution:  Notice that one of the binomials has a + instead of the – that is in the standard form of the circle.  Rewrite the addition as subtraction.

y+7 is the same as y-(-7).

The other binomial doesn’t have any number added or subtracted.  We can rewrite this by subtracting zero.

  x is the same as x-0.

The rewritten equation is as follows.

(x-0)^2+(y-(-7))^2=12

Then, if you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=0.  In the binomial with y, the number after the minus sign is k.  Thus, k=-7.  The r^2 lines up with 25.  So r^2=12 which means r=sqrt{12}=2sqrt{3}.

This is the equation of a circle with a center of (0, -7) and a radius of r=2sqrt{3} .

Graphing an Absolute Value Equation Shifted Horizontally by Plotting Points

Plotting Points Method

  1. Find ordered pair solutions
    • Choose values for one variable
    • Find the value for the other variable
  2. Plot the ordered pair solutions
  3. Draw the line or curve that connects the ordered pairs

Example: Graph  y=delim{|}{x-2}{|} by the plotting points method.

1. Find ordered pair solutions.  We can organize this information in a table.

  • Choose values for one variable

When the equation is written in “y=” format it is easier to choose values for x and then find the y’s.  Although you are allowed to choose any x values, for this table I have used a standard set of x’s so that we have a variety.  Some negatives, some positives and zeros.

xy
-2
-1
0
1
2
  • Find the values for the other variable

To find the y’s substitute the value of x into the equation and simplify to find y.

xyyy
-2|-2-2|=|-4|=4
-1|-1-2|=|-3|=3
0|0-2|=|-2|=2
1|1-2|=|-1|=1
2|2-2|=|0|=0

2.  Plot the ordered pair solutions

Using the table above we have 5 ordered pairs (-2, 4), (-1, 3), (0, 2), (1, 1), (2, 0)

Plot the ordered pairs using the rectangular coordinate system.

3.  Draw the line or curve that connects the ordered pairs.

When you look at the plotted ordered pairs you should see a pattern the that points make.  As I discussed in a previous post, most absolute value graphs form a “v” shape. It seems I am not graphing the portion of the graph that shows the “v” shape.  With the plotting points method all I can do is find more ordered pairs to try to get the correct graph.  However, in a future post, we will discuss other graphing methods give us more information about where the interesting part of the graph is.

In the mean time, to finish this graph I need just one more ordered pair.  (3, 1) which I get from plugging three into the equation for x.

This makes the graph have the “v” shape we expected.

The line represents all of the ordered pairs that are solutions to the equation y=delim{|}{x-2}{|}.

 

Graphing an Absolute Value Equation by Plotting Points

Plotting Points Method

  1. Find ordered pair solutions
    • Choose values for one variable
    • Find the value for the other variable
  2. Plot the ordered pair solutions
  3. Draw the line or curve that connects the ordered pairs

Example: Graph  y=delim{|}{x}{|}+2 by the plotting points method.

1. Find ordered pair solutions.  We can organize this information in a table.

  • Choose values for one variable

When the equation is written in “y=” format it is easier to choose values for x and then find the y’s.  Although you are allowed to choose any x values, for this table I have used a standard set of x’s so that we have a variety.  Some negatives, some positives and zeros.

xy
-2
-1
0
1
2
  • Find the values for the other variable

To find the y’s substitute the value of x into the equation and simplify to find y.

xyyy
-2|-2|+2=2+2=4
-1|-1|+2=1+2=3
0|0|+2=0+2=2
1|1|+2=1+2=3
2|2|+2=2+2=4

2.  Plot the ordered pair solutions

Using the table above we have 5 ordered pairs (-2, 4), (-1, 3), (0, 2), (1, 3), (2, 4)

Plot the ordered pairs using the rectangular coordinate system.

3.  Draw the line or curve that connects the ordered pairs.

When you look at the plotted ordered pairs you should see a pattern the that points make.  In this case, the points form a “v” shaped graph. Most of the equations with absolute value have graphs of this shape.

The line represents all of the ordered pairs that are solutions to the equation y=delim{|}{x}{|}+2.

Graphing a Quadratic Equation by Plotting Points

Plotting Points Method

  1. Find ordered pair solutions
    • Choose values for one variable
    • Find the value for the other variable
  2. Plot the ordered pair solutions
  3. Draw the line or curve that connects the ordered pairs

Example: Graph  y= x^2-5 by the plotting points method.

1. Find ordered pair solutions.  We can organize this information in a table.

  • Choose values for one variable

When the equation is written in “y=” format it is easier to choose values for x and then find the y’s.  Although you are allowed to choose any x values, for this table I have used a standard set of x’s so that we have a variety.  Some negatives, some positives and zeros.

xy
-2
-1
0
1
2
  • Find the values for the other variable

To find the y’s substitute the value of x into the equation and simplify to find y.

xy
-2(-2)^2-5=4-5=-1
-1(-1)^2-5=1-5=-4
0(0)^2-5=0-5=-5
1(1)^2-5=1-5=-4
2(2)^2-5=4-5=-1

2.  Plot the ordered pair solutions

Using the table above we have 5 ordered pairs (-2, -1), (-1, -4), (0, -5), (1, -4), (2, -1)

Plot the ordered pairs using the rectangular coordinate system.

3.  Draw the line or curve that connects the ordered pairs.

When you look at the plotted ordered pairs you should see a pattern the that points make.  In this case, the points form a “u” shaped curve called a parabola. Most of the equations with x squared have graphs of this shape.

The line represents all of the ordered pairs that are solutions to the equation y= x^2-5.

Here is a youtube video with a similar example.

Graphing a Linear Equation by Plotting Points (m is a fraction)

Plotting Points Method

  1. Find ordered pair solutions
    • Choose values for one variable
    • Find the value for the other variable
  2. Plot the ordered pair solutions
  3. Draw the line or curve that connects the ordered pairs

Example: Graph  y= -{1}/{3}x+2 by the plotting points method.

1. Find ordered pair solutions.  We can organize this information in a table.

  • Choose values for one variable

When the equation is written in “y=” format it is easier to choose values for x and then find the y’s.  For this table I will choose multiples of the denominator so that my ordered pairs are all integers.

xy
3(-2)=-6
3(-1)=-3
3(0)=0
3(1)=3
3(2)=6
  • Find the values for the other variable

To find the y’s substitute the value of x into the equation and simplify to find y.

xyyy
-6-1/3(-6)+2=2+2=4
-3-1/3(-3)+2=1+2=3
0-1/3(0)+2=0+2=2
3-1/3(3)+2=-1+2=1
6-1/3(6)+2=-2+2=0

2.  Plot the ordered pair solutions

Using the table above we have 5 ordered pairs (-6, 4), (-3, 3), (0, 2), (3, 1), (6, 0)

Plot the ordered pairs using the rectangular coordinate system.

3.  Draw the line or curve that connects the ordered pairs.

When you look at the plotted ordered pairs you should see a pattern the that points make.  In this case, the points form a straight line.

The line represents all of the ordered pairs that are solutions to the equation y= - {1} / {3}x+2.

Here is a youtube video with a similar example.

Graphing a Linear Equation by Plotting Points (m is an integer)

Plotting Points Method

  1. Find ordered pair solutions
    • Choose values for one variable
    • Find the value for the other variable
  2. Plot the ordered pair solutions
  3. Draw the line or curve that connects the ordered pairs

 

Example: Graph  y= 2x-5 by the plotting points method.

1. Find ordered pair solutions.  We can organize this information in a table.

  • Choose values for one variable

When the equation is written in “y=” format it is easier to choose values for x and then find the y’s.  Although you are allowed to choose any x values, for this table I have used a standard set of x’s so that we have a variety.  Some negatives, some positives and zeros.

xy
-2
-1
0
1
2
  • Find the values for the other variable

To find the y’s substitute the value of x into the equation and simplify to find y.

xy
-22(-2)-5=-4-5=-9
-12(-1)-5=-2-5=-7
02(0)-5=0-5=-5
12(1)-5=2-5=-3
22(2)-5=4-5=-1

2.  Plot the ordered pair solutions

Using the table above we have 5 ordered pairs (-2, -9), (-1, -7), (0, -5), (1, -3), (2, -1)

Plot the ordered pairs using the rectangular coordinate system.

3.  Draw the line or curve that connects the ordered pairs.

When you look at the plotted ordered pairs you should see a pattern the that points make.  In this case, the points form a straight line.

The line represents all of the ordered pairs that are solutions to the equation y= 2x-5.

Here is a youtube video with a similar example