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.Com
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) (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 we can see that the equation is in slope-intercept form where the slope is and the y-intercept is
Since 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
Also, because the slope is , this means:
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
So this is the graph of through the points
and
=================================================================================
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.
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