Algebra 2
- XC Sabreur
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Algebra 2
Can someone please explain to me how to graph an equation? For example, one of the homework problems is " y= -2x-3. So how would i graph that? If anyone can help it is appreciated. Thank you
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And if anyone else here fences or runs cross country or track, PM me ! we should talk sometime, I love running and fencing
- Emily
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Re: Algebra 2
Do you not have a graphing calculator?
I highly recommend you get one.. You'll use it tons and tons in Alg. 2. TI 83 or something.
I highly recommend you get one.. You'll use it tons and tons in Alg. 2. TI 83 or something.
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Re: Algebra 2
oooh just found the best site for you!!! It graphs stuff for you. gOOD LUCK! CLICK!
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Re: Algebra 2
Wow, that site is sweet. Thanks a bunch, I see what to do now. Its cool, thanks again.
I'm a moderator of the Family/Friends forum, and Relationships! If you ever need help with anything, feel free to email, IM, or PM me.
For those of you wondering what my name means, here it is. The XC stands for Cross Country, because I run cross country and I run distance in track for my school. And Sabreur because I fence sabre in my spare time. And no, its not a lightsabre.
And if anyone else here fences or runs cross country or track, PM me ! we should talk sometime, I love running and fencing
For those of you wondering what my name means, here it is. The XC stands for Cross Country, because I run cross country and I run distance in track for my school. And Sabreur because I fence sabre in my spare time. And no, its not a lightsabre.
And if anyone else here fences or runs cross country or track, PM me ! we should talk sometime, I love running and fencing
- junglemonkey
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Re: Algebra 2
Well although that site may be good, I figured I'd explain it since you won't have it in an exam
y= -2x - 3 is in the form y = mx + c so therefore the graph is a straight line.
In the form y = mx + c,
m = gradient
c = y-intercept
The gradient is the steepness of the slope of the line right? and that's vertical height/horizontal distance.
The y-intercept is where the line crosses the y axis
So, if the equation was y = 3x - 4
m = 3
c = -4
Anyway, in your case, m = -2 and c = -3
So (if you're using graph paper you can be more accurate) plot the point (0, -3) (on the y axis x = 0) on the y axis for the intercept.
The line would slope from the top left to bottom right since the gradient is negative.
And in this case m=-2. So, that's -2/1 ... OR ... That's vertical height over horizontal distance, do you know how to plot that?
you could also find another point on the line and join the two rather than going through plotting the gradient
Find out where it crosses the x axis. On the x axis y=0 so substitute that in your formula.
Apologies for any mistakes i made!
y= -2x - 3 is in the form y = mx + c so therefore the graph is a straight line.
In the form y = mx + c,
m = gradient
c = y-intercept
The gradient is the steepness of the slope of the line right? and that's vertical height/horizontal distance.
The y-intercept is where the line crosses the y axis
So, if the equation was y = 3x - 4
m = 3
c = -4
Anyway, in your case, m = -2 and c = -3
So (if you're using graph paper you can be more accurate) plot the point (0, -3) (on the y axis x = 0) on the y axis for the intercept.
The line would slope from the top left to bottom right since the gradient is negative.
And in this case m=-2. So, that's -2/1 ... OR ... That's vertical height over horizontal distance, do you know how to plot that?
you could also find another point on the line and join the two rather than going through plotting the gradient
Find out where it crosses the x axis. On the x axis y=0 so substitute that in your formula.
Apologies for any mistakes i made!
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Re: Algebra 2
I thought it was y=mx+b...mabye i'm just dumb...i doubt it matters anyways...
Have you ever tried to step in my shoes?
Have you ever tried to balance on that beam?
And if you ever tried to step in my shoes
They'll never be quite as soft as they seem.
-incubus
Have you ever tried to balance on that beam?
And if you ever tried to step in my shoes
They'll never be quite as soft as they seem.
-incubus
- junglemonkey
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Re: Algebra 2
Yeah same thing. so you use b as the y intercept??
we use the equation in the form of y = mx + c
we use the equation in the form of y = mx + c
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- She
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Re: Algebra 2
y=mx+b
This is slope-intercept form. m is the slope of the line (rise over run) and b is the point where the line intercepts the y axis (called the y-intercept).
You'll also see things in standard form:
Ax+By=C
To get it into slope intercept form so you can plot it: -A/B (the opposite of A divided by = m (the slope) and C/B (C divided by =b (the small b in slope-intercept form--B and b are not the same in these).
So if you have the equasion 2x+3y=5
2 divided by 3 is 2/3 (fraction), and the opposite of that is a negative so m=-2/3 (also your slope).
5 divided by 3 is 5/3 (fraction), so that's 5/3=b (the y-intercept).
In slope intercept form that's:
y=-2/3x+5/3
So on your coordinate grid you find the Y axis (up and down) and put a dot on it at positive 5/3. Then count down two whole numbers, and to the right three whole numbers. Put another dot, and from there count down another two and right another three. Then draw a line connecting the points and extending beyond with little arrows at the ends.
Hope this helps. If you need any more help, PM me or whatnot.
This is slope-intercept form. m is the slope of the line (rise over run) and b is the point where the line intercepts the y axis (called the y-intercept).
You'll also see things in standard form:
Ax+By=C
To get it into slope intercept form so you can plot it: -A/B (the opposite of A divided by = m (the slope) and C/B (C divided by =b (the small b in slope-intercept form--B and b are not the same in these).
So if you have the equasion 2x+3y=5
2 divided by 3 is 2/3 (fraction), and the opposite of that is a negative so m=-2/3 (also your slope).
5 divided by 3 is 5/3 (fraction), so that's 5/3=b (the y-intercept).
In slope intercept form that's:
y=-2/3x+5/3
So on your coordinate grid you find the Y axis (up and down) and put a dot on it at positive 5/3. Then count down two whole numbers, and to the right three whole numbers. Put another dot, and from there count down another two and right another three. Then draw a line connecting the points and extending beyond with little arrows at the ends.
Hope this helps. If you need any more help, PM me or whatnot.
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- ToraToraTora
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Re: Algebra 2
well, in calc if you ever take it, you wont be allowed to use a graphing calculator. But something you should know is:
y=m(x-a)+b
if you have any variables in the a spot, it will affect how the graph moves left t right. If the thing says (x-1), it moves right 1 spot. And positive moves left. the slope is m, so you should be able to make a decent graph.
y=m(x-a)+b
if you have any variables in the a spot, it will affect how the graph moves left t right. If the thing says (x-1), it moves right 1 spot. And positive moves left. the slope is m, so you should be able to make a decent graph.
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