Petenerd

06-10-2009, 10:36 PM

I need help finding the y intercepts of these points.

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

View Full Version : [MERGED] Y-intercept of six points

Petenerd

06-10-2009, 10:36 PM

I need help finding the y intercepts of these points.

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

Petenerd

06-10-2009, 10:38 PM

I need help finding the y intercepts of these points.

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

thanks

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

thanks

Loren

06-10-2009, 10:39 PM

If the graph of an equation crosses the y-axis, it has a y-intercept. Points don't have y-intercepts.

Petenerd

06-10-2009, 10:48 PM

Is the y-intercept where the line touches the y-axis?

Mrspi

06-10-2009, 11:00 PM

I need help finding the y intercepts of these points.

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

thanks

You COULD graph each of these lines, and try to determine FROM THE GRAPH where the line crosses the y-axis (the y-intercept). But graphing is tedious and notoriously inaccurate, lots of times.

So...I'll help you with the first one. Let's write the equation of the line through the two points (-1, 3) and (1, -1).

Start by finding the slope. The slope of the line through (x[sub:3eenav3l]1[/sub:3eenav3l], y[sub:3eenav3l]1[/sub:3eenav3l]) and (x[sub:3eenav3l]2[/sub:3eenav3l], y[sub:3eenav3l]2[/sub:3eenav3l]) can be found using this formula, where "m" is the slope:

m = (y[sub:3eenav3l]2[/sub:3eenav3l] - y[sub:3eenav3l]1[/sub:3eenav3l])/(x[sub:3eenav3l]2[/sub:3eenav3l] - x[sub:3eenav3l]1[/sub:3eenav3l])

m = (-1 - 3) / [1 - (-1)]

m = (-4) / (2)

m = -2

Now, use the slope-intercept form of the equation of a line,

y = mx + b

"m" is the slope, and "b" is the y-intercept.

We know the slope is -2. Substitute -2 for "m":

y = -2x + b

We now need to find "b". Since we know that the point (1, -1) is on the line, the equation must be true when x = 1 and y = -1. Substitute 1 for x and -1 for y:

y = -2x + b

-1 = -2(1) + b

-1 = -2 + b

Solve this for "b". Add 2 to both sides of the equation:

-1 + 2 = -2 + b + 2

1 = b

So, the slope-intercept form for the equation of this line is

y = -2x + 1

OH...you wanted the y-intercept for this line, right?? WELL..."b" is the y-intercept...and that is 1!

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

thanks

You COULD graph each of these lines, and try to determine FROM THE GRAPH where the line crosses the y-axis (the y-intercept). But graphing is tedious and notoriously inaccurate, lots of times.

So...I'll help you with the first one. Let's write the equation of the line through the two points (-1, 3) and (1, -1).

Start by finding the slope. The slope of the line through (x[sub:3eenav3l]1[/sub:3eenav3l], y[sub:3eenav3l]1[/sub:3eenav3l]) and (x[sub:3eenav3l]2[/sub:3eenav3l], y[sub:3eenav3l]2[/sub:3eenav3l]) can be found using this formula, where "m" is the slope:

m = (y[sub:3eenav3l]2[/sub:3eenav3l] - y[sub:3eenav3l]1[/sub:3eenav3l])/(x[sub:3eenav3l]2[/sub:3eenav3l] - x[sub:3eenav3l]1[/sub:3eenav3l])

m = (-1 - 3) / [1 - (-1)]

m = (-4) / (2)

m = -2

Now, use the slope-intercept form of the equation of a line,

y = mx + b

"m" is the slope, and "b" is the y-intercept.

We know the slope is -2. Substitute -2 for "m":

y = -2x + b

We now need to find "b". Since we know that the point (1, -1) is on the line, the equation must be true when x = 1 and y = -1. Substitute 1 for x and -1 for y:

y = -2x + b

-1 = -2(1) + b

-1 = -2 + b

Solve this for "b". Add 2 to both sides of the equation:

-1 + 2 = -2 + b + 2

1 = b

So, the slope-intercept form for the equation of this line is

y = -2x + 1

OH...you wanted the y-intercept for this line, right?? WELL..."b" is the y-intercept...and that is 1!

Loren

06-10-2009, 11:49 PM

Yes.

Subhotosh Khan

06-11-2009, 07:17 AM

I need help finding the y intercepts of these points.

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

As noted before, probably you wanted to find the y-intercepts of the lines joining the pair of points.

To do that algebraically, you need to find the equation of the straight-line joining two points (x[sub:qqsaorsr]1[/sub:qqsaorsr], y[sub:qqsaorsr]1[/sub:qqsaorsr]) and (x[sub:qqsaorsr]2[/sub:qqsaorsr], y[sub:qqsaorsr]2[/sub:qqsaorsr]). That equation is:

(y - y[sub:qqsaorsr]1[/sub:qqsaorsr])/(y[sub:qqsaorsr]2[/sub:qqsaorsr] - y[sub:qqsaorsr]1[/sub:qqsaorsr]) = (x - x[sub:qqsaorsr]1[/sub:qqsaorsr])/(x[sub:qqsaorsr]2[/sub:qqsaorsr] - x[sub:qqsaorsr]1[/sub:qqsaorsr])

now set x = 0 and find y = y-intercept

(-1,3), (1,-1)

(-6,2), (0,4)

(0,3), (5,0)

Thanks

As noted before, probably you wanted to find the y-intercepts of the lines joining the pair of points.

To do that algebraically, you need to find the equation of the straight-line joining two points (x[sub:qqsaorsr]1[/sub:qqsaorsr], y[sub:qqsaorsr]1[/sub:qqsaorsr]) and (x[sub:qqsaorsr]2[/sub:qqsaorsr], y[sub:qqsaorsr]2[/sub:qqsaorsr]). That equation is:

(y - y[sub:qqsaorsr]1[/sub:qqsaorsr])/(y[sub:qqsaorsr]2[/sub:qqsaorsr] - y[sub:qqsaorsr]1[/sub:qqsaorsr]) = (x - x[sub:qqsaorsr]1[/sub:qqsaorsr])/(x[sub:qqsaorsr]2[/sub:qqsaorsr] - x[sub:qqsaorsr]1[/sub:qqsaorsr])

now set x = 0 and find y = y-intercept

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