Question

Use the suggested method to sketch the graph of each linear relationship.
a) y= -4x+3 (slope, y-intercept method)
b) 2x-y=4 (x-,y-intercept method)
c) 3x=6y-3 (table of values)

Answer 1

a.) y = -4x + 3 --- slope, y int method..
In y = mx + b form, the slope is in the m position and the y intercept is in the b position.
So...the slope is -4 (same as -4/1) and the y int is 3, or (0,3).

First we plot out y int points (0,3).....and since our slope is -4/1, we come down 4 units (because of the negative), and go right 1 unit, then down 4 units, the to the right 1..
If you want to make sure you are doing it right, find the x int by subbing in 0 for y.
y = -4x + 3
0 = -4x + 3
4x = 3
x = 3/4....so the x int is (3/4,0)
And following the slope, coming down from (0,3).....coming down 4 and to the right 1, you should cross the x axis at (3/4,0)

Looks like this ..

http://www.meta-calculator.com/online/?panel-102-graph&data-bounds-xMin=-10&data-bounds-xMax=10&data-bounds-yMin=-7.28&data-bounds-yMax=7.28&data-equations-0=%22y%20%3D%20-4*x%20%2B%203%22&data-rand=undefined&data-hideGrid=false

any questions about this one ?

Answer 2

b.) 2x - y = 4 ....x and y intercept method

To find x intercept, sub in 0 for y
2x - 0 = 4
2x = 4
x = 2.....so the x intercept is (2,0)

To find the y intercept, sub in 0 for x
2(0) - y = 4
-y = 4
y = -4 ...so the y intercept is (0,-4)

so now you plot your points (2,0) and (0,-4)......connect them with a solid line

Looks like this...
http://www.meta-calculator.com/online/?panel-102-graph&data-bounds-xMin=-10&data-bounds-xMax=10&data-bounds-yMin=-7.28&data-bounds-yMax=7.28&data-equations-0=%222*x%20-%20y%20%3D%204%22&data-rand=undefined&data-hideGrid=false

Answer 3

c.) 3x - 6y = 3.....now we will sub in numbers for x to find y

Lets let x = 0
3(0) - 6y = 3
-6y = 3
y = -1/2

Lets let x be 1
3(1) - 6y = 3
3 - 6y = 3
-6y = 3 - 3
-6y = 0
y = 0

Lets let x be 2
3(2) - 6y = 3
6 - 6y = 3
-6y = 3 - 6
-6y = -3
y = 1/2

one more...let x be 3
3(3) - 6y = 3
9 - 6y = 3
-6y = 3 - 9
-6y = -6
y = 1

so...out table now has 4 sets of points...

x | y
----------------
0 -1/2
1 0
2 1/2
3 1

Now it is just a matter of plotting your points (0,-1/2) , (1,0), (2,1/2) , (3,1)

Looks like this...
http://www.meta-calculator.com/online/?panel-102-graph&data-bounds-xMin=-10&data-bounds-xMax=10&data-bounds-yMin=-7.28&data-bounds-yMax=7.28&data-equations-0=%223*x%20-%206*y%20%3D%203%22&data-rand=undefined&data-hideGrid=false

Answer 4

@texaschic101 wow thank you so much