Posted by Amy on Tuesday, November 9, 2010 at 3:19pm.
Can someone please answer the questions I posted? Thank you.
Hello. I have worked on this and just need confirmation. 1. I need to find the intercepts and then use them to graph the equation. 2y-6=2x; I have x = 0 for y - intercept; 2y-6=2(0); 2y-6 = 0; 2y = 6; y = 3. I set y at 0 also. 2 (0) - 6 = 2x; -6=2x; -3=x. so the points are
(-3,0), (0,3). 2. Identify the y - intercept. I used the formula y=mx+b; y intercept is (0,3); 3.
Amy, all of your calculations are correct; but I thought you should have
rearranged your Eq first:
y = mx + b. b is the y- Int.
2y - 6 = 2x.
2y = 2x + 6,
Divide both sides by 2:
y = x + 3.
b = 3 = y- Int.
Let y = 0:
Then x = -3 = x-Int.
Coordinates:
(0 , 3) , (-3 , 0).
Yes! Thank you I have since worked it out and have gotten the answer you came up with. :)
The graph of equation is expanded vertically by a factor of 3 and reflected over the x-axis. how would you describe the function?
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