Graphing ordered pairs
The ordered pair (4, 5) lies in quadrant ..
Quadrant I
the points (3,1), (5,3) & (7,5) make straight line because
they the same slope
In y = mx + b. Each variable represents
Y= how far up/down
m= the slope, how steep the line is
x= how far long
b= y-intercept when x=0
when graphing y=-x+7 and y=2x+1 and seeing the lines cross at (2, 5), I conclude
the point ( 2, 5) is the solution of the system
y= 2x-5 has the following features
positive slope and negative y-intercept
To graph the ordered pair (-5,-2)on a plane I go...
Go back 5 and down 2
In the equation y= mx+b , m is... and b is...
m is the slope and b is the y-intercept
I can describe y=-3x in those terms
the slope is negative and y-intercept is zero and the line goes through the origin
Given y=-x-3 and y= x+ 1, graphing these lines will lead (0,-3),(0,1), or (-2,-1) as the solution of the system of equations
(-2,-1) as the solution to the system
Graphing y≥ x+3 will lead to shading the region
above y=x+3
The x-axis is the line on the graph that runs
Horizontally
y= 3x+ 2, find the coordinate of a point P when x=-2
P (-2,-4)
if I am graphing y≤ 3x+2, I will have a line
a line that is solid
one Fact of graphing a linear system in two is
The system is consistent and independent
The solution to a system of equations is the values of your variables that make
make all your equations true
The line passing through M(4,8) and N(5, Y). find y for MN to be parallel to the x-axis
y=8
the 4 symbols in linear inequalities are
less than, greater than, less than, or equal to, & greater than, or equal to
I am graphing y> -3, my line is... and I shade the region
my line is dotted, broken and I shade the region above the line y=-3
Graphing a dependent system of equations in two variables will lead
Infinitely many solutions
y= 2x+ 4, if x= -5, then y=
y=-6
the line that contains V, P, Q passing through x=2 is parallel to the y-axis. The x-coordinate for each point is
x=2
The cost to ride a taxi is $ 3 plus $ 2 per mile. A linear inequality that best represents this situation is
2x+3<=y
In y=2x+3, I change the slope to m=-2 and b=0, I describe the new line as
Negative line and passing through the origin
Cab A has a line y=2x+2 and Cab B has a line y=x+5, in the long run cab.... is offering the best deal
Cab B
An inconsistent system of linear equations in two has
No solution