How do you graph this?

Question

1. y=-2/3x-1

2. Finding the slope of the line that contains the points(3-5) and (5,-5)

3. Finding yhr equation of the line that contains the point (3,-5) and has slope 2/3

Answer 1

2.slope=(y2-y1)/(x2/x1)=0/2=0

3.equation of such a line will be y-y1=m(x-x1)
y+5=2/3(x-3)
3(y+5)=2(x-3)
3y+15=2x-6
so the equation is 2x-3y-21=0
in the standard form

1.give 3 values say -3,0 and 3 to x and find the corresponding values of y
y=1 or -1 and -3 respectively
now the ordered pairs are (-3,1),(0,-1),(3,-3)
plot these points and join them and produce on either side and you have the graph of your line

Answer 2

use slope-intercept form of a line: y = mx + b
m = slope of the line = rise/run = (y1-y2)/(x1-x2)
b = y-intercept...where the line will hit the y-axis
x & y = (x,y) = point on the line

1.) y = mx + b = -2/3x -1 = (-2/3)x + (-1), so
m = -2/3 & b = -1
Since b = -1, it will hit the y-axis at (0,-1), so that is your 1st point. From there, you use the slope. A slope of (-2)/3 means you would go down 2 and right 3 from (0, -1)

2.) slope = m = rise/run = (y2-y1)/(x2-x1)
(x1, y1) = (3, -5) & (x2, y2) = (5, -5), so
((-5) - (-5))/(5 - 3) = (-5 + 5)/(5 - 3) = 0/2 = 0
The slope is 0, so it is a horizontal line at y = -5

3.) y = mx + b, m = 2/3, (x1,y1) = (3,-5), b = ?
In order to find b plug in m, x, & y
-5 = (2/3)(3) + b
b = -5 - 2 = -7
y = mx + b = (2/3)x - 7
y-int: (0, -7)
From that point use the slope and go up 2 and right 3 to find the next point.

Answer 3

I'm not here to do your homework, ya know...

Answer 4

TRY GOING TO GOOGLE