Ch. 1.2 Graphing Equations
A point (x,y) are on the graph of the equation if substituting those values
for x and y still render a true statement (that is, satisfy the equation).
Example 1
Determine if the following points are on the graph of the equation 2x-y =6
a) (2,3)
2(2) – 3 =? 6
4-3 = 1 ≠ 6
No (2,3) is not on the graph.
b) (2,-2)
2(2) – (-2) = 4 + 2= 6
6=6
Yes (2,-2) is on the graph
p.19 #16
If (2,b) is a point on the graph of y = x2 + 3x, what is b?
Substitute 2 for x and b for y.
b = 22 + 3(2) = 4 + 6 = 10
Example 2
Graph the equation y = 2x + 5
Because this equation is in the form y = mx + b, we know it’s a line. To
graph the equation, we need to know at least two points on the line.
When x = 0, y = 2(0) + 5 = 5
When x = 2, y =2(2) + 5 = 9
10
9
8
7
6
x
0
2
y
5
9
5
4
3
2
1
0
-1 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1 0
0
-2
-3
-4
-5
1
2
3
4
5
6
7
8
9 10
Ch. 1.2
p.14 USING A GRAPHING CALCULATOR TO GRAPH EQUATIONS
To graph an equation in two variables, x and y, using a graphing
calculator requres that the equation be written tin the form y=
{expression in x}. If the original equation is not in this form, there are
ways to obtain an equivalent equation in the “y=“ form.
Procedures “Allowed” in order to maintain equivalent
equations:
1. You can switch sides of the equation and it remains the
same: Ex: 3x + 5 = y is the same is y = 3x + 5
2. You can simplify the dies of an equation by combining like
terms, eliminating parentheses, and son on:
Ex: 2y + 2 + 6 = 2x + 5(x +1) is the same as
2y + 8 = 7x + 5
3. You can add or subtract the same expression on both sides
of the equation.
Ex: y + 3x – 5 = 4 is the same as
y + 3x – 5 + 5 = 4 + 5, which can be simplified to
y + 3x = 9
4. You can multiply or divide both sides of the equation by the
same nonzero expression:
Ex:
3y = 6 – 2x can be multiplied on both sides by 1/3
1
1
• 3 y = (6 − 2 x ) which can be simplified to
3
3
2
y = 2− x
3
SETTING THE STANDARD VIEWING WINDOW
The viewing window is the portion of the coordinate plane
visible on your graphing calculator screen. It is defined by
Xmin, Xmax, Ymin, and Ymax. Xscl and Yscl define the
distance between tick marks on the x-axis and y-axis,
respectively.
To display the current window variable values, press the
WINDOW button.
The standard values for the viewing window are
Xmin = -10
Xmax = 10
Xscl = 1
Ymin = -10
Ymax = 10
Yscl = 1
These standard values can be automatically set by
pressing the ZOOM button and then selecting
6:Zstandard.
Example 5 p. 15
Graph the equation: 6x2 + 3y = 36
Step 1: Use Equivalent Equations Procedures to solve for y in
terms of x.
Subtract 6x2 from both sides:
3y = -6x2 + 36
Divide both sides by 3:
y = - 2x2 + 12
What is the y-intercept of this equation?
y=12
So we should reset the Ymax in the viewing window to be at least
12.
To graph this, press the Y= button and type in the expression.
Press (-)
2
X,T, θ
x2
+
1
2
Then press GRAPH
To reproduce this graph manually on your own graph paper,
it is helpful to create a table of x and y values for the equation.
This can be done with the Table function in your calculator.
Press 2nd
[TBLSET] to display the TABLE SETUP screen.
Press (-)
1 0
Press 1
Press
ENTER
ENTER
ENTER
to set TblStart = -10
to set ∆Tbl = 1.
ENTER
To automatically generate independent and dependent values.
to display the table screen.
Press 2nd TABLE
INTERCEPTS
The points, if any, at which a graph crosses or touches the coordinate axes are called the
intercepts. The x-coordinate of a point at which the graph crosses or touches the x-axis
is an x-intercept. The y-coordinate of a point at which the graphs crosses or touches the
y-axis is a y-intercept. Because the x-intercepts of the graph of an equation are those xvalues for which y=0, they are also called the zeros (or roots) of the equation.
14
12
10
8
6
4
x-intercepts
2
(-2,0)
-10
-8
-6
-4
(2,0)
0
-2
0
2
4
6
8
10
-2
-4
-6
y-intercept
(0,-4)
PROCEDURE FOR FINDING INTERCEPTS
1. To find the x-intercept(s), if any, of the graph of an equation, let y=0 in the
equation and solve for x.
2. To find the y-intercepts(s), if any, of the graph of an equation let x=- in the
equation and solve for y.
x-intercepts of y=x2 -4
Set y=0
0 = x2 – 4
4 = x2
x2 = 4
x = +/- 2
And remember y=0, so
x-intercepts are
(-2,0) and (2,0)
y-intercepts of y=x2 -4
Set x=0
y = 02 – 4
y = -4
And remember x=0, so
y-intercept is
(0,-4)
What about y = x2 + 4?
24
22
20
We can see that
the y-intercept is
(0,4), but what
are the xintercepts for this
graph?
18
16
14
12
10
8
6
4
2
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Set y=0 to find x-intercepts.
0 = x2 + 4
-4 = x2
x2= -4
x=
± −4
This is not a real number, so there are no
x-intercepts for this equation.
1.3 SYMMETRY; GRAPHING KEY EQUATIONS; CIRCLES
Symmetry
Knowing an equation’s symmetry can help you know how to graph it.
A graph can be symmetric with respect to the x-axis, for every
point (x,y) on the graph, the point (x, -y) is also on the graph.
If you fold the graph at the x-axis graphs line up to each other.
A graph can be symmetric with respect to the y-axis, for every
point (x,y) on the graph, the point (-x, y) is also on the graph.
If you fold the graph at the y-axis graphs line up to each other.
A graph can be symmetric with respect to the origin, for every
point (x,y) on the graph, the point (-x, -y) is also on the graph.
If you fold the graph at x-axis and then at the y-axis graphs line up to each other.
This is the graph of y = x2
It is symmetric to the y-axis.
To check, plug in (-x,y) and see if you get an equivalent equation.
y = (-x)2 = x2
Therefore it is symmetric to the y-axis
90
70
50
30
10
-50
-40
-30
-20
- 10
0
10
20
30
40
50
- 10
-30
-50
This is the graph of x = y2
It is symmetric to the x-axis
To check, plug in (x,-y) and see if you get an equivalent equation.
x = (-y)2 = y2
Therefore it is symmetric to the y-axis
4
3
2
1
0
-10
-8
-6
-4
-2
0
-1
-2
-3
-4
2
4
6
8
10
This is the graph of y = x3
It is symmetric to the origin.
To check, plug in (-x,-y) and see if you get an equivalent equation.
-y = (-x)3 = -x3 [since the exponent, 3, is an odd number]
Multiply both sides by -1 and you get y=x3.
Therefore it is symmetric to the origin.
1000
800
600
400
200
0
-60
-50
-40
-30
-20
-10
0
-200
-400
-600
-800
-1000
10
20
30
40
50
60
x
y=x^2
y=x^3
-10
100
-1000
-9
81
-729
-8
64
-512
-7
49
-343
-6
36
-216
-5
25
-125
-4
16
-64
-3
9
-27
-2
4
-8
-1
1
-1
0
0
0
1
1
1
2
4
8
3
9
27
4
16
64
5
25
125
6
36
216
7
49
343
8
64
512
9
81
729
10
100
1000
x
y = sqrt(x)
y=-sqrt(x)
0
0
0
0.25
0.5
-0.5
0.5
0.707107
-0.70711
0.75
0.866025
-0.86603
1
1
-1
1.25
1.118034
-1.11803
1.5
1.224745
-1.22474
1.75
1.322876
-1.32288
2
1.414214
-1.41421
2.25
1.5
-1.5
2.5
1.581139
-1.58114
2.75
1.658312
-1.65831
3
1.732051
-1.73205
3.25
1.802776
-1.80278
3.5
1.870829
-1.87083
3.75
1.936492
-1.93649
4
2
-2
4.25
2.061553
-2.06155
4.5
2.12132
-2.12132
How to make Tables in Excel. Select your chosen x and y columns for the XY-Scatter CHART.
x
y=x^2
y=x^3
x
y = sqrt(x)
y=-sqrt(x)
-10
=A2^2
=A2^3
0
=SQRT(E2)
=-SQRT(E2)
-9
=A3^2
=A3^3
=E2+0.25
=SQRT(E3)
=-SQRT(E3)
-8
=A4^2
=A4^3
=E3+0.25
=SQRT(E4)
=-SQRT(E4)
-7
=A5^2
=A5^3
=E4+0.25
=SQRT(E5)
=-SQRT(E5)
-6
=A6^2
=A6^3
=E5+0.25
=SQRT(E6)
=-SQRT(E6)
-5
=A7^2
=A7^3
=E6+0.25
=SQRT(E7)
=-SQRT(E7)
-4
=A8^2
=A8^3
=E7+0.25
=SQRT(E8)
=-SQRT(E8)
-3
=A9^2
=A9^3
=E8+0.25
=SQRT(E9)
=-SQRT(E9)
-2
=A10^2
=A10^3
=E9+0.25
=SQRT(E10)
=-SQRT(E10)
-1
=A11^2
=A11^3
=E10+0.25
=SQRT(E11)
=-SQRT(E11)
0
=A12^2
=A12^3
=E11+0.25
=SQRT(E12)
=-SQRT(E12)
1
=A13^2
=A13^3
=E12+0.25
=SQRT(E13)
=-SQRT(E13)
2
=A14^2
=A14^3
=E13+0.25
=SQRT(E14)
=-SQRT(E14)
3
=A15^2
=A15^3
=E14+0.25
=SQRT(E15)
=-SQRT(E15)
4
=A16^2
=A16^3
=E15+0.25
=SQRT(E16)
=-SQRT(E16)
5
=A17^2
=A17^3
=E16+0.25
=SQRT(E17)
=-SQRT(E17)
1.3 CIRCLES
A circle is a set of oins in the XY-plane that are a fixed distance, r, from a
fixed point, (h,k). The distance, r, is the radius, and the point (h,k) is the
center.
We can use the distance formula to derive the equation for a circle.
Remember the distance formula --
DISTANCE FORMULA
The distance between two points P1 = (x1,y1) and P2 = (x2,y2) is
d(P1,P2) =
( x2 − x1 ) 2 + ( y2 − y1 ) 2
For every point (x,y) on the graph of a circle, the distance to (h,k) must
always be r.
r = ( x − h) 2 + ( y − k ) 2
For the standard equation form of a circle, square both sides and then
switch sides of the equation to get:
( x − h) 2 + ( y − k ) 2 = r 2
What if the center, (h,k), is (0,0)?
Then the equation becomes ( x − 0) 2
+ ( y − 0) 2 = r 2
so
x2 + y2 = r 2
A unit circle is a circle with center (0,0) and radius = 1. The equation for a
unit circle is:
x2 + y2 = 1
Example 6 p. 27
Radius = 5, center is (-3,6). Write the standard equation.
(x − (−3) )2 + ( y − 6) 2 = 52
( x + 3) 2 + ( y − 6) 2 = 25
Now do #39 on p.31
Example 7
Graph this equation: ( x + 3) 2 + ( y − 2) 2 = 16
In standard form this equation becomes:
(x − (−3) )2 + ( y − 2) 2 = 42
From this we can tell that the center is (-3,2) and the radius is 4.
7
6
5
4
3
r=4
(-3,2)
2
1
0
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
-5
-6
1
2
OLD, LONG WAY to GRAPH A CIRCLE ON A CALCULATOR:
To graph a circle with a graphing calculator you have to get y by itself
in the form “y= “. (It’s more trouble than it’s worth!)
First, put all terms with x and constant expressions on one side:
( y − 2) 2 = 16 − ( x + 3) 2
Take the square root of both sides. Remember the answer is +/-
( y − 2) = ± 16 − ( x + 3) 2
add 2 to both sides
y = 2 ± 16 − ( x + 3) 2
On your calculator, use the Y = editor
Y1 = 2 + 16 − ( x + 3) 2
Y2 = 2 − 16 − ( x + 3) 2
Knowing that the center is (-3,2) and the radius is 4 will help you with
setting the viewing window.
You will get a distorted view with the standard viewing window.
It will look like an oval.
You will need a Square Screen to get a normal view.
Press
, then select 6:ZSquare, then press
ZOOM
GRAPH
Now do #57 on p. 31
NEW, EASY WAY to GRAPH A CIRCLE ON A CALCULATOR:
You can also generate a circle from the DRAW function.
To display the DRAW menu, press 2nd [DRAW]. Select 9:Circle(
Example 8 on p. 29
Graph the equation x2+y2 + 4x – 6y+12 = 0
What is the center and what is the radius?
We need to put this equation in STANDARD form [(x-h)2 + (y-k)2=r2] to
easily see the center and radius. We do this completing the square. Group
the terms involving x, then group the terms involving y, then put the
constants on the right hand side.
(x2 + 4x + 4) + (y2-6y + 9) = - 12 + 4 + 9
Complete the square of the x’s by halving the coefficient of x and squaring
it. (4/2)2 = 4. Make sure you also add this on the right side.
Complete the square of the y’s by halving the coefficient of y and squaring
it (6/2)2 = 9. Make sure you also add this on the right side.
(x+2)2 + (y-3)2 =1
In standard form this is
(x-(-2))2 + (y-3)2 = 12
So from this we can see that it is a circle with center (-2,3) and radius=1
5
4
3
r=1
(-2,3)
2
1
0
-5
-4
-3
-2
-1
0
-1
1
Example 9
Find the general form of a circle, x2 + y2 + ax + by + c = 0,
which has a center of (1,-2) and contains the point (4,-2).
Notice that we are not given the radius,r. However, we know
that r is the distance from any point on the circle to the center.
So
r = (4 − 1) 2 + (− 2 − (− 2))
2
r = 32 − 0
r = 9 =3
We can generate the general form by using the standard form of the circle equation.
( x − 1) 2 + ( y + 2) 2 = 9
Multiply out and rearrange
x2 − 2x +1+ y 2 + 4 y + 4 = 9
Combine like terms, then subtract 9 from both sides to get 0 on the right side.
x2 + y2 − 2x + 4 y + 5 − 9 = 9 − 9
x2 + y2 − 2x + 4 y − 4 = 0
GENERAL FORM OF CIRCLE EQUATION
HOMEWORK #2
p.19 1.2 Exercises
#1,9,13,15,19,25,39
p.30 1.3 Exercises
#3,13,17,29,31,41,43,47,53, 59, 63