• Oct 13th 2009, 03:50 PM
BsMummy
This question may be a bit vague but considering i cant draw the graph i hope someone can still help

I need to find the gradient and y-intercept of a line given, than i need to write the equation of the line.

Now am i correct in saying the y-intercept is where the line crosses on the y point?

If so than it crosses at -1

so how do i than find the gradient if that is all im given?
• Oct 13th 2009, 04:10 PM
pickslides
Quote:

Originally Posted by BsMummy

Now am i correct in saying the y-intercept is where the line crosses on the y point?

Let's call it the y-axis, do you have the point where your line crosses the x-axis?

• Oct 13th 2009, 04:11 PM
Quote:

Originally Posted by BsMummy
This question may be a bit vague but considering i cant draw the graph i hope someone can still help

I need to find the gradient and y-intercept of a line given, than i need to write the equation of the line.

Now am i correct in saying the y-intercept is where the line crosses on the y point?

If so than it crosses at -1

so how do i than find the gradient if that is all im given?

Yes, you are correct about the y-intercept. In order to find the gradient of a line you need two points. If you know the y-intercept, then you have the point (0,-1). But you have to have another point to find the gradient. What other information are you given? There must be more. Are you given a graph of the line?
• Oct 13th 2009, 04:18 PM
BsMummy
Quote:

Originally Posted by pickslides
Let's call it the y-axis, do you have the point where your line crosses the x-axis?

The line crosses the x-axis at 4
• Oct 13th 2009, 04:19 PM
BsMummy
Quote:

Yes, you are correct about the y-intercept. In order to find the gradient of a line you need two points. If you know the y-intercept, then you have the point (0,-1). But you have to have another point to find the gradient. What other information are you given? There must be more. Are you given a graph of the line?

would finding the x-axis help? if so that would be 4

so now i have (0,-1) and (0,4)

is that all i will need to find the gradient?

yes i am given a graph of the line but i dont know how to but that graph up on here!
• Oct 13th 2009, 04:24 PM
pickslides
Quote:

Originally Posted by BsMummy

so now i have (0,-1) and (0,4)

No, you really have (0,-1) and (4,0)

the gradient $m = \frac{y_2-y_1}{x_2-x_1}$

where $(0,-1) = (x_1,y_1)$ and $(4,0) = (x_2,y_2)$
• Oct 13th 2009, 04:59 PM
BsMummy
okay so i have worked out that the gradient is .5 or 1/2 - but i worked that out by looking at the graph and counting how many sqaures i went up from the y intercept to the x intercept

so now that i have the gradient at .5 or 1/2 how do i write the equation of the line?
• Oct 13th 2009, 05:04 PM
Quote:

Originally Posted by BsMummy
okay so i have worked out that the gradient is .5 or 1/2 - but i worked that out by looking at the graph and counting how many sqaures i went up from the y intercept to the x intercept

so now that i have the gradient at .5 or 1/2 how do i write the equation of the line?

The general form is $y=mx+b$. So the line will have the equation $y=\frac{1}{2}x+b$ where b is the y-intercept. Since b=-1, the line is $y=\frac{1}{2}x-1$
• Oct 13th 2009, 05:10 PM
BsMummy
okay i used the way to find the gradient that pickslides gave and now im getting 1/4 - which im thinking would be the right answer since im no maths genious!
• Oct 13th 2009, 05:11 PM
BsMummy
Quote:

The general form is $y=mx+b$. So the line will have the equation $y=\frac{1}{2}x+b$ where b is the y-intercept. Since b=-1, the line is $y=\frac{1}{2}x-1$