Question 206143: how do i graph the lines by using equations from x(-3,3) and how do i determine the gradient of y-intercept, showing the domain and range in eachcase A y=2x+1
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! The question is not specific, but reading between the lines, let's solve what we know
.
If base eqn is,,,,,y=2x+1,,,,slope is 2, and y intercept is 1
.
a line parallel to this also has slope of 2.
.
to define y intercept, insert point ( -3,3 ) into eqn with same slope
.
y= 2x +b,,,,,(3) =2(-3) +b ,,,,,,b=9 (y intercept)
.
or new eqn is ,,,,,y=2x+9
.
a line perpendicular to original, has slope = negative reciprocal
.
y= -1/2 x +b,,,,inserting (-3,3),,,(3) = (-1/2)(-3) +b,,,,and b= 1.5
.
or new eqn is y=-1/2 x +1.5
.
The Domain and Range of all lines is (- infinity, + infinity)
.
or ALL REAL NUMBERS
.
Points of theory are: y = mx +b, with m=slope, b= y intercept
.
parallel lines have m1=m2, but perpendicular lines have m2=-1/m1
.
to consider a point, insert in equation and solve for b
.
Hopefully this helps, if not pls resubmit problem thanks
.
|
|
|
