kuta software infinite algebra 1 graphing lines

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Graphing Linear Equations; Part 1 · (Algebra) Number Lines Part 1

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Question:Hey, I can't seem to get graphing perpendicular lines on the Cartesian Graph. I understand that the slope-intercept formula for the second line must be the reciprocal and the opposite positive or negative sign than the original line, but when I get past this point, I am not able to lengthen the line to pass through the correct Y-axis intercept... How can I get it to pass through the right Y-axis intercept?? Hope i explained it enough, if not, just ask, thanks! :) Catbek: Yes, I understand that completly, but even when I have that part 100% correct I STILL don't get it to pass through the correct Y-intercept... :/ Ummm... my formula is: y=mx -/+ b

Answers:So you have the gradient and two points I am assuming. Use y - y1 = m(x - x1) to find the equation of the line. Then substitute in 0 for x into the equation to find the y intercept as when the line passes through the y intercept, x = 0.

Question:I have to graph two equations, and write the solution. for coincident lines, my teacher said you write either "all points fall on x = #" or "all points equal x = #" but I forget which one it is. which one is it, or are they both right?

Answers:I don't understand "all points fall on x = #" or "all points equal x = #" but for coincident lines, the equations are equivalent -- can be shown to be the same equation. The solution points for both equations fall on the same line.

Question:Line 1: -x -7y +9 = 0 Line 2: 4x+3y+14 =0 Line 1, Line 2, both lines or neither? 2,1 -5,2 7,5 I have to make a choice, if Line one is at 2,1, -5,2, 7,5, same with Line 2, or both lines at the same place or neither lines at none of these choices

Answers:Yes, both are lines. .

Question:I don't know how to do this and th etecher didn't even teach us. It's not in our books either:// Please dont just give me an answer, I do my homework. Bu tplease tell me why or how you get an answer. THANKS problem: graph by plotting: y=x^2 y=x-1 Can you describe how to graph/solve.

Answers:1) y=x^2 The easiest way to do these is to make a table of the values, then graph the points. For this one since x is squared, you would consider both the positive and negative of the same number -- i.e. you would get the same value for y. For example, x=3 and x= -3, both give y = 9 x ------ y ---------------- points to graph 0 ----- 0 ---------------- (0,0) +- 1/2 ----- 1/4 ------------- (1/2,1/4) (-1/2, 1/4) +- 1 ------ 1 -------------- (1,1 ) (-1,1) +- 2 ------- 4 -------------- (2,4) (-2,4) +- 3 ----- 9 -------------- (3,9) (-3,9) You can do a few more x values, but this gives you an idea. ------------------------------------------------------- 2) y=x-1 Please read over the reference as it is one of the better ones I've found to explain y=mx+b, where m is the slope (rise over run) and b is the y-intercept. The y-intercept is just the point on the y axis when x is 0. So the y-intercept is -1 or the point (0, -1). The slope is 1. 1 times x is just x. You can think of it a 1/1, and go up one and over one -- to (1,0). Or you can think of it as -1 / -1 and go down one and back one -- to (-1, -2). Again you could make a table with x and y values and get these points x -------- y --------- point -2 -------- -3 ---------- (-2, -3) -1 --------- -2 --------- (-1, -2) 0 ------- -1 -------- (0, -1) 1 --------- 0 ----------- (1, 0) 2 --------- 1 ----------- (2, 1)

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Algebra: graphing lines 1 :Graphing linear equations

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