Thread: Slope-Intercept

I have to get this weekend homework done quick! My parents come home in about 6 hours and it HAS to be done before that! I'm suppose to "Write the slope-intercept form of an equation of the line that passes through the given point and is parallel to the graph of each equation." I really don't understand the y=mx+b formula and I know it has a big part in the problems. I'm given a line and the slope-intercept equation of it and I think my teacher said all it was for is to take the slope from it and then use the two coordinates given (-2,-3) to plug in and have a line parallel to the first one. If someone can explain to me where the numbers go in the formula, I should be able to work the other problems. Here is #1: y=2x-1 and the given point is (-2,-3). So where does 2,-2,and -3 go in the y=mx=b formula??

I have to get this weekend homework done quick! My parents come home in about 6 hours and it HAS to be done before that! I'm suppose to "Write the slope-intercept form of an equation of the line that passes through the given point and is parallel to the graph of each equation." I really don't understand the y=mx+b formula and I know it has a big part in the problems. I'm given a line and the slope-intercept equation of it and I think my teacher said all it was for is to take the slope from it and then use the two coordinates given (-2,-3) to plug in and have a line parallel to the first one. If someone can explain to me where the numbers go in the formula, I should be able to work the other problems. Here is #1: y=2x-1 and the given point is (-2,-3). So where does 2,-2,and -3 go in the y=mx=b formula??

You need more tools in your bag of tricks.

I really don't understand the y=mx+b formula

Then I think you're sunk. You need this. If you don't get this one, how are you going to understand the "Point-Slope" form, the "2-Point" form, the "Standard Form", or the "Intercept Form"?

You say you have a teacher. That's a good start. Have you also a book? There MUST be some background information in there.

Yes I have a teacher and a book. But when she taught this, I didn't get it and then she assigned weekend homework. If you could just tell me where those numbers plug into the formula, then I could work the rest of them. The sheet has an equation and a line on a graph representing it. Then there is a point to the left of it and it's position is (-2,-3). I have to figure out the equation for a line parallel to the first using the slope of the first one and the coordinates of the point. So the three numbers I have are: 2 -the slope of the first equation, and (-2,-3) which are the coordinates of the point. I have to plug these three numbers into y=mx=b and that is the answer. I just don't know where each number goes. I think I remember my teacher saying you plug a number into y,m,x, and then b remains as it is without being replaced by a number. So where do the three numbers go in the formula?

Originally Posted by leedownen

I have to get this weekend homework done quick! My parents come home in about 6 hours and it HAS to be done before that! I'm suppose to "Write the slope-intercept form of an equation of the line that passes through the given point and is parallel to the graph of each equation." I really don't understand the y=mx+b formula and I know it has a big part in the problems. I'm given a line and the slope-intercept equation of it and I think my teacher said all it was for is to take the slope from it and then use the two coordinates given (-2,-3) to plug in and have a line parallel to the first one. If someone can explain to me where the numbers go in the formula, I should be able to work the other problems. Here is #1: y=2x-1 and the given point is (-2,-3). So where does 2,-2,and -3 go in the y=mx=b formula??

Let your line be y = mx + c. It has to be parallel to the line y = 2x - 1. But parallel lines have the SAME gradient and the gradient of y = 2x - 1 is 2. Therefore m = 2.

So your line becomes y = 2x + c. How do we get c??? Well, you know that (-2,-3) is a point on your line, that is, when y = -3, x = -2:

-3 = 2(-2) + c.

So you can solve for c: c = 1.

Then your line becomes y = 2x + 1 and that's the answer.

One more question! When you solved for c, the problem was -3=2(-2)+c! how do you solve all those numbers to equal 1??

edit: (for the second time!) Oh! I think I get it. Is this correct?
-3=2(-2)+c
-3=-4+C
move 4 to right side of equal sign and it the problem is 1=c

Originally Posted by leedownen

move 4 to right side of equal sign and it the problem is 1=c

You MUST step up your game a little, here. It is not acceptable to "move" things. Try addition.

-3 = -4 + c

-3 + 4 = -4 + 4 + c

1 = 0 + c = c

No kidding, if you stick to more fundamentally correct behavior, you will learn more, you will be prepared better for the future, and you will be more consistent.