SOLUTION: For problems 4-6. Graph each equation. State the slope, y-intercept, and x-intercept for each 4.) -4x+y = -4 5.)-2x-3y=-6 6.)x-4=-9 For problems 7-8, construct a table

Algebra ->  Graphs -> SOLUTION: For problems 4-6. Graph each equation. State the slope, y-intercept, and x-intercept for each 4.) -4x+y = -4 5.)-2x-3y=-6 6.)x-4=-9 For problems 7-8, construct a table       Log On


   

Question 942168: For problems 4-6. Graph each equation. State the slope, y-intercept, and x-intercept for each
4.) -4x+y = -4
5.)-2x-3y=-6
6.)x-4=-9
For problems 7-8, construct a table of at least three ordered pairs and graph each equation.
7.)y=x cubed-3
8.)y=|x-4|
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
4.
Solved by pluggable solver: Graphing Linear Equations


-4%2Ax%2B1%2Ay=-4Start with the given equation



1%2Ay=-4%2B4%2Ax Add 4%2Ax to both sides

y=%281%29%28-4%2B4%2Ax%29 Multiply both sides by 1

y=%281%29%28-4%29-%281%29%28-4%29x%29 Distribute 1

y=-4%2B%284%29x Multiply

y=4%2Ax-4 Rearrange the terms

y=4%2Ax-4 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=4 (the slope) and b=-4 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-1

y=4%2A%28-1%29-4

y=-4-4 Multiply

y=-8 Add

So here's one point (-1,-8)





Now lets find another point

Plug in x=0

y=4%2A%280%29-4

y=0-4 Multiply

y=-4 Add

So here's another point (0,-4). Add this to our graph





Now draw a line through these points

So this is the graph of y=4%2Ax-4 through the points (-1,-8) and (0,-4)


So from the graph we can see that the slope is 4%2F1 (which tells us that in order to go from point to point we have to start at one point and go up 4 units and to the right 1 units to get to the next point) the y-intercept is (0,-4)and the x-intercept is (1,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=-4 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,-4).


So we have one point (0,-4)






Now since the slope is 4%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,-4), we can go up 4 units


and to the right 1 units to get to our next point



Now draw a line through those points to graph y=4%2Ax-4


So this is the graph of y=4%2Ax-4 through the points (0,-4) and (1,0)



5.
Solved by pluggable solver: Graphing Linear Equations


-2%2Ax-3%2Ay=-6Start with the given equation



-3%2Ay=-6%2B2%2Ax Add 2%2Ax to both sides

y=%28-1%2F3%29%28-6%2B2%2Ax%29 Multiply both sides by -1%2F3

y=%28-1%2F3%29%28-6%29%2B%281%2F3%29%28-2%29x%29 Distribute -1%2F3

y=6%2F3-%282%2F3%29x Multiply

y=%28-2%2F3%29%2Ax%2B6%2F3 Rearrange the terms

y=%28-2%2F3%29%2Ax%2B2 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-2%2F3 (the slope) and b=2 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-9

y=%28-2%2F3%29%2A%28-9%29%2B2

y=18%2F3%2B2 Multiply

y=24%2F3 Add

y=8 Reduce

So here's one point (-9,8)





Now lets find another point

Plug in x=-6

y=%28-2%2F3%29%2A%28-6%29%2B2

y=12%2F3%2B2 Multiply

y=18%2F3 Add

y=6 Reduce

So here's another point (-6,6). Add this to our graph





Now draw a line through these points

So this is the graph of y=%28-2%2F3%29%2Ax%2B2 through the points (-9,8) and (-6,6)


So from the graph we can see that the slope is -2%2F3 (which tells us that in order to go from point to point we have to start at one point and go down -2 units and to the right 3 units to get to the next point), the y-intercept is (0,2)and the x-intercept is (3,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=2 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,2).


So we have one point (0,2)






Now since the slope is -2%2F3, this means that in order to go from point to point we can use the slope to do so. So starting at (0,2), we can go down 2 units


and to the right 3 units to get to our next point



Now draw a line through those points to graph y=%28-2%2F3%29%2Ax%2B2


So this is the graph of y=%28-2%2F3%29%2Ax%2B2 through the points (0,2) and (3,0)


6.
Solved by pluggable solver: Graphing Linear Equations


1%2Ax-4%2Ay=-9Start with the given equation



-4%2Ay=-9-1%2Ax Subtract 1%2Ax from both sides

y=%28-1%2F4%29%28-9-1%2Ax%29 Multiply both sides by -1%2F4

y=%28-1%2F4%29%28-9%29%2B%281%2F4%29%281%29x%29 Distribute -1%2F4

y=9%2F4%2B%281%2F4%29x Multiply

y=%281%2F4%29%2Ax%2B9%2F4 Rearrange the terms

y=%281%2F4%29%2Ax%2B9%2F4 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=1%2F4 (the slope) and b=9%2F4 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-9

y=%281%2F4%29%2A%28-9%29%2B9%2F4

y=-9%2F4%2B9%2F4 Multiply

y=0%2F4 Add

y=0 Reduce

So here's one point (-9,0)





Now lets find another point

Plug in x=-5

y=%281%2F4%29%2A%28-5%29%2B9%2F4

y=-5%2F4%2B9%2F4 Multiply

y=4%2F4 Add

y=1 Reduce

So here's another point (-5,1). Add this to our graph





Now draw a line through these points

So this is the graph of y=%281%2F4%29%2Ax%2B9%2F4 through the points (-9,0) and (-5,1)


So from the graph we can see that the slope is 1%2F4 (which tells us that in order to go from point to point we have to start at one point and go up 1 units and to the right 4 units to get to the next point) the y-intercept is (0,2.25) ,or (0,9%2F4), and the x-intercept is (-9,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=9%2F4 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,9%2F4).


So we have one point (0,9%2F4)






Now since the slope is 1%2F4, this means that in order to go from point to point we can use the slope to do so. So starting at (0,9%2F4), we can go up 1 units


and to the right 4 units to get to our next point



Now draw a line through those points to graph y=%281%2F4%29%2Ax%2B9%2F4


So this is the graph of y=%281%2F4%29%2Ax%2B9%2F4 through the points (0,2.25) and (4,3.25)



7.
y=x%5E3-3
x|y
0|-3.....y=0%5E3-3=>y=-3
1|-2....y=1%5E3-3=>y=-2
-1|-4....y=%28-1%29%5E3-3=-1-3=>y=-4
-2|-11
2|5



8.
y=abs%28x-4%29

y=abs%28x-4%29
x|y
0|4
4|0
1|3
5|1
7|3