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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TMJ on February 08, 2013, 05:57:47 pm

Title: yr 11 methods gradient?
Post by: TMJ on February 08, 2013, 05:57:47 pm

Say you're asked to sketch -x-y=1, when rearranged in y=mx+c form you get, -y=x+1. What does the -y mean. Does it mean the y intercept becomes -1, or the gradient become -. Or do you find the x and y intercepts by substituting zero for each?

Title: Re: yr 11 methods gradient?
Post by: e^1 on February 08, 2013, 05:59:00 pm

Quote from: TMJ on February 08, 2013, 05:57:47 pm

Say you're asked to sketch -x-y=1, when rearranged in y=mx+c form you get, -y=x+1. What does the -y mean. Does it mean the y intercept becomes -1, or the gradient become -. Or do you find the x and y intercepts by substituting zero for each?

Just multiply both sides by -1 so that you can get it in the form y = mx + c.

Title: Re: yr 11 methods gradient?
Post by: abcdqdxD on February 08, 2013, 06:29:20 pm

Quote from: e^1 on February 08, 2013, 05:59:00 pm

Just multiply both sides by -1 so that you can get it in the form y = mx + c.

Yeah, so you would end up with:

y=-x-1, which means gradient is -1 and y intercept is -1

Title: Re: yr 11 methods gradient?
Post by: Yacoubb on February 08, 2013, 06:32:12 pm

-x-y=1

1) Rearrange the equation into the y = mx + c form.

-y = x + 1
y = -x -1

2) Your gradient then becomes the coefficient of x.

Hence, -1 is your gradient.

Start at your y intercept, (0,-1), move down one unit and 1 to the side (follow the path denoted by the gradient).

Alternatively, you could use the intercept method where you find the x-intercept and y-intercept, and then plot the points and join these points by a straight line.

Title: Re: yr 11 methods gradient?
Post by: LazyZombie on February 08, 2013, 07:16:50 pm

BTW Class of 2014! :D

(I don't want the thread to die xD)