SOLUTION: I need help graphing the equation: y-3=3(x+1) And plotting the points: (1, -3) and y-int -3, and x-int 1 I read the rules i promise this is for the same problem. And thank

Algebra ->  Linear-equations -> SOLUTION: I need help graphing the equation: y-3=3(x+1) And plotting the points: (1, -3) and y-int -3, and x-int 1 I read the rules i promise this is for the same problem. And thank       Log On


   

Question 763665: I need help graphing the equation: y-3=3(x+1)
And plotting the points: (1, -3) and y-int -3, and x-int 1
I read the rules i promise this is for the same problem. And thank you in advance!

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First solve for y and get it into slope intercept form y = mx+b

y-3=3(x+1)

y-3=3x+3

y=3x+3+3

y=3x+6

Looking at y=3x%2B6 we can see that the equation is in slope-intercept form y=mx%2Bb where the slope is m=3 and the y-intercept is b=6


Since b=6 this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point




Now since the slope is comprised of the "rise" over the "run" this means
slope=rise%2Frun

Also, because the slope is 3, this means:

rise%2Frun=3%2F1


which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3 and over 1



So starting at , go up 3 units


and to the right 1 unit to get to the next point



Now draw a line through these points to graph y=3x%2B6

So this is the graph of y=3x%2B6 through the points and






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The second problem doesn't make any sense. If you had the points (1,-3) and (0,-3), which is the y-intercept, then you would have a horizontal line. This horizontal line has NO x-intercept. So there has to be a typo somewhere.