Graphing
Consider two points C(h,k) and P(x,y) ;and the distance between them is r. From the distance formula:
square both sides
r^{2}= ( x - h )^{2}+ ( y - k )^{2}
r^{2}= ( x - h )^{2 }+ ( y - k )^{2} is the standard equation for circles.
Ax^{2}+ Ay^{2}+ Dx + Ey + F = 0
is the general form of the equation for circles.
ex.
Write the general equation of the circle with center at C(4,-5) and a
radius of 5.
r^{2}= ( x - h )^{2}+ ( y - k )^{2}
5^{2}= ( x - 4 )^{2}+ ( y -(-5))^{2}
25 = x^{2} -8x + 16 + y^{2} + 10y + 25
x^{2}+ y^{2 }-8x + 10y + 16 = 0 is the general equation.
Find the center and the radius of the circle with equation x^{2}+ y^{2}-10x - 4y + 16 = 0.
x^{2}+ y^{2}-10x - 4y +16 = 0
(x-10x ) + (y- 4y ) = -16 complete the square.
(x^{2}-10x + 25) + (y^{2}- 4y + 4) = -16 + 25 + 4
( x -5)^{2 }+ ( y - 2 )^{2} = 13
The center is at (5, 2) and the radius is
If the constant term in the standard equation is positive, then a graph of a circle exist. If it is zero, a single point exist, If it is negative, there is no graph.
standard form of the equation of an ellipse:
where b^{2}= a^{2} - c^{2}
center (0,0) (0,0)major axis on x-axis, length 2a on y-axis, length 2aminor axis on y-axis, length 2b on x-axis, length 2bfoci (Â±c,0) (0,Â±c)vertices (Â±a,0) (0,Â±a)covertices (0,Â±b) (Â±b,0)</span>
a is always larger than b; and a,b, and c are related by c^{2}= a^{2} - b^{2}
ex.
Graph 9x^{2}+ 4y^{2}= 36
which is an ellipse.
a^{2} = 9 ; b^{2} = 4
center (0,0)
major axis: y-axis
vertices: (0, Â± 3)
covertices: (Â± 2,0)
Graph 9x^{2 }+ 25y^{2 }= 225
a^{2}= 25 ; b^{2}= 9
center (0,0)
major axis: x-axis
vertices: (Â± 5,0)
covertices (0,Â± 3)
c^{2}= a^{2} - b^{2}
c^{2}= 25 - 9
c^{2} = 16
foci: (Â± 4,0)
Hyperbola:
The standard equation for hyperbolas is:
where b^{2} = c^{2} - a^{2}
vertices (Â± a,0) (0, Â± a)foci (Â± c,0) (0, Â± c)transverseaxis on x-axis, on y-axis,length 2a length 2aconjugateaxis on y-axis, on x-axis,length 2b length 2b
a is always larger than b; and a,b, and c are related by c^{2} = a^{2} + b^{2}
ex.
Graph 9x^{2} - 16y^{2}= 144
a^{2} = 16 ; b^{2} = 9
major axis: x-axis
vertices: (Â± 4,0)
c^{2} = a+ b
c^{2} = 16 + 9
c^{2} = 25
foci: (Â± 5,0)
Graph 36x^{2 }- 4y^{2 }+ 144 = 0
36x^{2} - 4y^{2} = -144 factor -1 out
4y^{2 }- 36x^{2} = 144
a^{2 }= 36 ; b^{2 }= 4
major axis: y-axis
vertices: (0,Â± 6)
c^{2}= a^{2 }+ b^{2}
c^{2}= 36 + 4
c^{2} = 40
(to find the asymptotes, let the x term equal the y term and solve for y)
Standard Equation: x<sup>2</sup>= 4py y<sup>2</sup>= 4px
vertical parabola horizontal parabola
axis y-axis x-axis
vertex (0,0) (0,0)
p is the focus of p>0 opens up p>0 opens right
the parabola
p<0 opens down p<0 opens left
focus (0,p) (p,0)
directrix y = -p x = -p
length of latus rectum 4|p| 4|p|
ex.
Graph x^{2}= 6y
The standard equation is x^{2 }= 4py,
which implies a vertical parabola with the vertex at the origin and the axis is the y-axis.
4p = 6
P > 0, the parabola opens up and the focus is at (0, 3/2)
The directrix is the line y = -3/2
4|p| = 6, the end points of the latus rectum is (3, 3/2) and (-3, 3/2)
Graph y^{2}= -16y
The equation is in the form: y^{2 }= 4px,
which implies a horizontal parabola with the vertex at the origin and the axis is the x-axis.
4p = -16
p = -4
p < 0, the parabola opens left and the focus is at ( -4,0).
The directrix is the line x = 4.
Â± 2p = Â± 8; \ the end points of the latus rectum are points (-4, 8) and (-4, -8).
Coordinate Systems and Graphing
Consider two number lines, one horizontal and one vertical. The two lines are perpendicular to each other at there zero points. These number lines make up the axes of a coordinate system. The horizontal line is called the x-axis, which is positive to the right and negative to the left. The vertical line is called the y-axis, which is positive going up and negative going down. The intersection of both x-axis and y-axis at there zero points is called the origin. The intersection of the axes also divides the plane into four regions called quadrants. The quadrants are numbered counterclockwise form one to four.
Ordered Pairs are pairs of real numbers that describe a point in the coordinate system. The ordered pair describes a point in the plane by its x and y coordinates. (x,y) is a point in the plane. The first number describes the distance in the x-axis; also called the abscissa, and the second number describes the distance in y-axis, also called the ordinate. The point (0,0) is the origin of the graph. Points are named by using capital letters, and point O often refers to the origin (0,0). (x,y) is the coordinate of the point.
Each quadrant have a distinct pattern;
In the first quadrant, both the x and y-coordinate are positive.
In the second quadrant, the x-coordinate is negative and the y-coordinate is positive.
In the third quadrant, both the x and y-coordinate are negative.
In the fourth quadrant, the x-coordinate is positive and the y-coordinate is negative.
Distance Between Points:
Distance Formula:
ex.
Find the distance between a(3,6) and B(7,9).
Graphing Linear Equations and Inequalities:
A solution for an equation with two variables is an ordered pair of real numbers that satisfy the equation. Consider the equation y = x + 3; for every number that is substituted into x a number y is produced and an ordered pair is formed.
Notice that there is an unlimited number of solutions to the equation.
This is the basis for plotting points or graphing an equation or inequality. Linear Equations have the basic form of Ax + By = C.
Graphing Linear Equations:
x and y-intercepts:
x-intercepts of a graph are the x-coordinates that are common with the x-axis.
To solve for the x-intercept, let y = 0, then solve for x.
y-intercepts of a graph are the y-coordinates that are common with the y-axis.
To solve for the y-intercept, let x = 0, then solve for y.
ex.
graph 2x -3y = 6
First, find the intercepts.
x = 0; -3y = 6
y=2; the point (0,-2) is on the line.y = 0; 2x = 6x = 3; the point (3,0) is on the line.
Now, check for a third point in the equation,
Let x = 6; 2(6) -3y = 612 -3y = 6-3y = -6y = 2
Graph y = 3x
x = 0; y = 3(0) = 0
The origin is part of the graph, and another two points must be found.
A table of values would be useful for this.
Vertical and Horizontal Lines:
x = 4, y = -5 are examples of vertical and horizontal equations of lines.
Graphing Linear Inequalities:
Graphing linear inequalities is almost the same as graphing linear equations, but with a slight difference.
ex.
graph 2x + 3y < 6
First graph the line 2x + 3y = 6, and use a broken line to graph the line since the line is not really part of the solution.
Choose a test point that is above and below the line to see if that point is a valid solution to the inequality. In this case, choose the origin (0,0) to see if it is a valid point in the solution.
2(0) + 3(0) < 60 < 6 the origin is part of thesolution.
Now, try a point above the line. Try point ( 5,4 ).
2(5) + 3(4) < 610 + 12 < 6 the point (5,4) is not part of the solution.
\ the solution graph is:
Any point in the blue area is a solution to the inequality.
Finding the Equation of a Line in the form Ax + B = C:
ex.
Find the equation of a line with a slope of -2 and a point at (4,6).
It is also known that a point (x,y) is on the line then,
From the formula:
Find the equation of the line that has the points (2,3) and (-5, 8)
First find the slope of the line.
Now that the slope is defined, the equation of the line can be found.
From the slope of the line, one of the given points and the variable point on the line (x,y); the equation of the line is from the formula;
Find the equation of the line with a slope of 3/2 and a y-intercept at 4.
The y-intercept implies that point (0,4) is in the line.
Now just solve as before:
3( x - 0 ) = 2( y - 4 )
3x = 2y - 8
3x - 2y = -8
To Translate the Graph:
ex.
(y - k) = | ( x - h) |
Replace (x - h) with x' and (y - k) with y':
Graph the equation with x' and y' with (h,k) as the center or vertex:
y' = | x' | at (h,k)
Slope-Intercept Form
Now, given a slope m and y-intercept at b.
What is the equation of the line?
y-intercept at b implies that the point (0,b) is on the line.
From the point slope form;
y - b = m ( x - 0 )
y - b = mx
y = mx + b This is the Slope-Intercept Form of a line.
ex.
From the previous example.
Find the equation of the line with a slope of 3/2
and a y-intercept at 4.
y-intercept at implies that the point (0,4) is on the line.
The slope-intercept form:
y = mx + b
2y = 3x + 8 multiplied 2 to both sides
3x - 2y = -8
Point-Slope Form:
The point-slope form is derived from the formula;
The equation of a line with point (x,y) and slope m is
THis is the point-slope form of a line.
ex.
From the previous example:
The equation of a line with slope -2 and the point (4,6)
y - 6 = -2( x - 4)
y - 6 = -2x + 8
2x + y = 14
If P_{1}(x_{1}, y_{1})and P_{2}(x_{2}, y_{2})are two different points, then the slope of the line between them is;
If the slope of a line is zero (0) then the line is a horizontal line.
If the slope of a line is undefined, then the line is a vertical line.
ex.
Find the slope of the line, m, between A(-3,1) and B(4,6)
Slope-Intercept Form
Now, given a slope m and y-intercept at b.
What is the equation of the line?
y-intercept at b implies that the point (0,b) is on the line.
From the point slope form;
y - b = m ( x - 0 )
y - b = mx
y = mx + b This is the Slope-Intercept Form of a line.
ex.
From the previous example.
Find the equation of the line with a slope of 3/2
and a y-intercept at 4.
y-intercept at implies that the point (0,4) is on the line.
The slope-intercept form:
y = mx + b
2y = 3x + 8 multiplied 2 to both sides
3x - 2y = -8
Y-axis symmetry:
The graph of an equation is symmetric with respect to the y-axis if an equivalent equation is obtained when x is replaced by -x.
ex.
x^{2} = 2y
(-x)^{2 }= 2y = x^{2 }= 2y
this equation is symmetric about the y-axis.
X-axis symmetry:
The graph of an equation is symmetric with respect to the x-axis if an equivalent equation is obtained when y is replaced by -y.
ex.
y^{6}= 4x
(-y)^{6 }= 4x = y^{6} = 4x
this equation is symmetric about the x-axis.
Origin Symmetry:
The graph of an equation is symmetric with respect to the origin if an equivalent equation is obtained when x is replaced by -x and y replaced by -y.
ex.
x^{3 }= y
(-x)^{3} = -y
-x^{3} = -y = x^{3} = y
this equation is symmetric about the origin.
Conic Sections Translation of Axes:
Ellipse:
standard equations for ellipse:
where b^{2} = a^{2} - c^{2}
center: (h,k) (h,k)major axis: x = h, length 2a y = k, length 2aminor axis: y = k, length 2b x = h, length 2bfoci: (h Â± c,k) (h, k Â± c)vertices: (h Â± a,k) (h, k Â± a)covertices: (h, k Â± b) (h Â± b,k)
a is always larger than b; and a,b, and c are related by c^{2 }= a^{2} - b^{2}
ex.
graph 16x^{2 }+ 25y^{2 }- 64x-200y + 64 = 0
convert to standard form
16x^{2 }- 64x + 25y^{2 }-200y = - 64
complete the square for the x and y terms
(16x^{2 }- 64x ) + (25y^{2 }-200y ) = - 64
16(x^{2 }- 4x ) + 25(y^{2 }- 8y ) = -64 complete the square
16(x^{2}- 4x + 4) + 25(y^{2}- 8y + 16) = -64 + 64 + 400
16(x - 2)^{2} + 25(y - 4)^{2} = 400
a = 5; b = 4
center: (2,4)
major axis: x = 2, length 10
minor axis: y = 4, length 8
c^{2 }= a^{2 }- b^{2}
c^{2} = 25 - 16
c^{2} = 9
c = 3
foci: (2 Â± 3,4) (5,4) and (-1,4)vertices: (2 Â± 5,4) (7,4) and (-3,4)covertices: (2, 4 Â± 4) (2,8) and (2,0)
ex.
graph 25x^{2 }+ 9y^{2}+ 200x + 54y + 256 = 0
convert to standard form
25x^{2 }+200x + 9y^{2}+ 4y = -256
complete the square for the x and y terms
(25x^{2}+200x )+ (9y^{2}+ 54y ) = - 256
25(x^{2 }+ 8x ) + 9(y^{2 }+ 6y ) = -256
25(x^{2} + 8x + 16) + 9(y^{2 }+ 6y + 9) = -256 + 400 + 81
25(x + 4)^{2} + 9(y + 3)^{2} = 225
a = 5; b = 3
center: (-4,-3)
major axis: y = -3, length 10
minor axis: x = -4, length 6
c^{2 }= a^{2 }- b^{2}
c^{2} = 25 - 9
c^{2} = 16
c = 4
foci: (-4, -3 Â± 4) (-4,1) and (-4,-7)vertices: (-4, -3 Â± 5) (-4,2) and (-4,-8)covertices: (-4 Â± 3,-3) (-1,-3) and (-7,-3)
Standard Equation for Hyperbolas:
where b^{2} = c^{2} - a^{2}
vertices (h Â± a,0) (0, k Â± a)foci (h Â± c,0) (0, k Â± c)</span>
a is always larger than b; and a,b, and c are related by c^{2} = a^{2} + b^{2}
ex.
Graph 9x<sup>2</sup> - 25y<sup>2 </sup>-54x + 250y -769 = 09x<sup>2 </sup>- 54x - 25y<sup>2 </sup>+ 250y = 769(9x<sup>2</sup> - 54x ) - ( 25y<sup>2 </sup>- 250y ) = 7699(x<sup>2 </sup>- 6x + 9) - 25(y<sup>2 </sup>- 10y + 25) = 769 +81 - 6259(x - 3)<sup>2</sup> - 25(y -5)<sup>2 </sup>= 225</span>
a = 5 ; b = 3
Center (3,5)
asymptotes
vertices (3 Â± 5,5)
ex.
16x^{2} - 9y^{2}- 224x - 54y + 847 = 0
16x^{2} - 224x -9y^{2} - 54y = -847
(16x^{2} - 224x ) - (9y^{2} - 54y ) = -847
16( x^{2} - 14x + 49) - 9( y^{2} + 6y + 9) = -847 +784 -81
16( x - 7)^{2} - 9(y + 3)^{2} = -144
9(y + 3)^{2} - 16( x - 7)^{2} = 144 Factor -1 out of both sides
a = 4; b = 3
Center (7,-3)
vertices (7,-3 Â± 4)
c^{2} = a^{2} + b^{2}
c^{2} = 16 + 9
c^{2} = 25
c = 5
foci (7,-3 Â± 5)
Conic Sections Translation of Axes:
Parabola:
The standard equation for translated parabolas is:
(x - h)<sup>2</sup> = 4p(y - k)<sup>2</sup> or (y - k)= 4p(x - h)vertex (h,k) (h,k)axis x = h y = kp is the focus of the parabola p > 0 opens up opens rightp < 0 opens down opens leftfocus (h, k + p) (h + p,k)directrix y = k - p x = h - plength of latus rectum 4|p| 4|p|endpoints of latus rectum (h Â± 2p,k + p) (h + p, k Â± 2p)
ex.
graph x<sup>2 </sup>- 6x - 8y + 49 = 0 convert to standard formx<sup>2 </sup>- 6x = 8y - 49 complete the square on the x termsx<sup>2 </sup>- 6x + 9 = 8y - 49 +9(x - 3)<sup>2</sup> = 8y - 40(x - 3)<sup>2</sup> = 8(y - 5) vertical parabola
vertex: (3,5)
axis: x = 3
4p = 8
p = 2 > 0; opens up
focus: (3,7)
directrix: y = 3
length of latus rectum: 4|2| = 8 (or Â± 2p)
endpoints of latus rectum: (7,7) and (-1,7)
ex.
graph y<sup>2 </sup>+ 8y + 12x - 8 = 0 convert to standard formy<sup>2 </sup>+ 8y = -12x + 8 complete the square on the x termsy<sup>2</sup> + 8y + 16 = -12x + 8 +16(y + 4)<sup>2</sup> = -12x + 24(y + 4)<sup>2</sup> = -12(x - 2) horizontal parabola
vertex: (2,-4)
axis: y = -4
4p = -12
p = -3 < 0; opens left
focus: (-1,-4)
directrix: x = 5
length of latus rectum: 4|-3| = 12 (or Â± 2p)
endpoints of latus rectum: (-1,2) and (-1,-10)
Links:
[1] http://www.course-notes.org/Subject/Math/Algebra