E Maths: Graph + Cube Root
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let's say you came across these questions:
a. Sketch the graphy of y= 2 (1-x) (x+2)
so what i would usually do is==>
y = 2 (1-x)(x+2)
y = 2 (x+2-x^2-2x)
y = 2 (-3x+2-x^2)
y = -6x + 4 - 2x^2 (then i take the numbers and times by -1)
y = 2x^2 + 6x -4
so if you're then asked to sketch the graph, HOW do you go about doing it? do you take x=-1,0,1,2,3... and substitute into the equation to get y and then plot the graph? seeming this is tedious, is there any other easier way?
b) Express x^2 - 6x + 7 in the form of (x-a)^2 + b and then sketch the graph of y=x^2 - 6x +7
so how would you approach this question???
this might sound silly especially at this time in time, but i'm not sure when qeustions ask for significant figures! example:
c) cube root [(0.57147 / 36783)]
so the 1st step i should do is to round up the numbers to 3 significant figures right?
0.57147 ~~> 0.571
36783 ~~> 36800
then the thing is, how would you find the answer to the cube root? usually i depend on the calculator for both square root and cube root, but since this is paper 1, how do you do them?
thanks everyone for helping out!
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a. Sketch the graphy of y= 2 (1-x) (x+2)
Don't go and expand
Notice that this is a negative x^2 curve
==> the final curve should be a frowning shape (n-shaped)if positive will be smiling shape
The curve will cut the x-axis at x=1 (when 1-x=0) and x = -2 (when x+2=0)
b) Express x^2 - 6x + 7 in the form of (x-a)^2 + b and then sketch the graph of y=x^2 - 6x +7
Complete the square (you should be able to do)
If cannot, get crucifixed or pillow to practice doing and post the solutionSketching wise, from (x-a)^2+b, since it is a positive x^2 curve, it is smiling curve (U-shaped)
Minimum point at x=a, y=b, i.e. point (a,b)
set (x-a)^2+b = 0
you get x = a ± sqrt (-b) => where the curve cuts the x-axis
====> if b is positive, curve does not cut x-axis at allc) cube root [(0.57147 / 36783)]
Are you taking the old syllabus? New syllabus paper 1 can use calculator
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Originally posted by gerlynn:
let's say you came across these questions:
a. Sketch the graphy of y= 2 (1-x) (x+2)
so what i would usually do is==>
y = 2 (1-x)(x+2)
y = 2 (x+2-x^2-2x)
y = 2 (-3x+2-x^2)
y = -6x + 4 - 2x^2 (then i take the numbers and times by -1)
y = 2x^2 + 6x -4
so if you're then asked to sketch the graph, HOW do you go about doing it? do you take x=-1,0,1,2,3... and substitute into the equation to get y and then plot the graph? seeming this is tedious, is there any other easier way?
b) Express x^2 - 6x + 7 in the form of (x-a)^2 + b and then sketch the graph of y=x^2 - 6x +7
so how would you approach this question???
this might sound silly especially at this time in time, but i'm not sure when qeustions ask for significant figures! example:
c) cube root [(0.57147 / 36783)]
so the 1st step i should do is to round up the numbers to 3 significant figures right?
0.57147 ~~> 0.571
36783 ~~> 36800
then the thing is, how would you find the answer to the cube root? usually i depend on the calculator for both square root and cube root, but since this is paper 1, how do you do them?
thanks everyone for helping out!
a. Sketch the graphy of y= 2 (1-x) (x+2)
If they ask you to sketch and you draw, you get 0. So NO plotting.
To sketch, you need the min/max point, x and y coordinate.
Express x^2 - 6x + 7 in the form of (x-a)^2 + b and then sketch the graph of y=x^2 - 6x +7
Eagle has got it.
c) cube root [(0.57147 / 36783)]
press the dam calculator.
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fyi davidche, mine is the old syllabus.
there's a difference btw sketch and draw?
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Originally posted by gerlynn:
there's a difference btw sketch and draw?
Of course there is, in paper 2, that's where you literally draw your graph, with every single point within the range given is plotted and your x and y axes are labelled so clearly and your graph is so proportioned.
In paper 1 most of the time, sketching the graph includes only a rough shape of the graph, where it cuts the x-axis (if it does) and where is the maximum or minimum point. :) and btw, you do not need to list your x-axis or y-axis to 1, 2, 3, 4 etc. Just indicate beside a point which is necessary (like if it's on the x-axis or if it is a maximum or minimum point) the coordinate of that point.
Nevertheless, your sketch of your graph must make sense, meaning you should know where your positive and negative coordinates are and your shape should be obvious. -
cube root [(0.57147 / 36783)
Er most likely without calculator like me the old syllabus they wont ask to go give the value of that cube root but likely to ask you to get a estimated figure of it
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thanks iamapebble!
eagle mentioned this:
Don't go and expand
Notice that this is a negative x^2 curve
==> the final curve should be a frowning shape (n-shaped)from the equation itself... how did he manage to spot it's a negative x^2? does this means it's a y = -x^2?
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Deleted. Wrong equations.
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Originally posted by gerlynn:
thanks iamapebble!
eagle mentioned this:
Don't go and expand
Notice that this is a negative x^2 curve
==> the final curve should be a frowning shape (n-shaped)from the equation itself... how did he manage to spot it's a negative x^2? does this means it's a y = -x^2?
Because from y= 2 (1-x) (x+2), we determine whether it's u or n shaped from the coefficient of x^2. From this equation, yes we don't go and expand it, but then we'd notice that (-x)*(x) = -x^2, therefore it is 'negative' which means it is a frowning shape as mentioned by eagle =)
Similary in any equations of a curve, we can determine the general shape from the coefficient of x^2, like if i say y = (x+1)(x-3), the coefficient of x^2 is definitely 1 which is positive, so it's a happy face, u shaped. =) -
Originally posted by gerlynn:
thanks iamapebble!
eagle mentioned this:
Don't go and expand
Notice that this is a negative x^2 curve
==> the final curve should be a frowning shape (n-shaped)from the equation itself... how did he manage to spot it's a negative x^2? does this means it's a y = -x^2?
By mentally expanding, negative of x multiplied by positive of x gives you negative of x squared. -
hi! sorry but i still dont get it :(
about the coefficient part and how to know which is which.
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Originally posted by gerlynn:
hi! sorry but i still dont get it :(
about the coefficient part and how to know which is which.
To determine the shape of a curve (since a straight line graph do not have many shapes), we always look at the coefficient x^2, because a general equation of a curve is ax^2 + bx + c. [It is due to the x^2 that we know it is a curve]
From the equation you gave, we don't expand it out, but then the only way we get the coefficient of x^2 from is only possible when you have x * x, and from y= 2 (1-x) (x+2) , you can only get your x^2 from (-x)*(x) = -x^2. If you expand it out you'll having
-x^2 + bx + c {noticing that bx + c do no affect your coefficient of x^2}.
Okay sounds confusing. o.o -
ooh! so that means for every equation we look for graphs, we have to look at x to determine shape of graph?
what about graphs for y=a/x? so we see the x to see if it's negative or positive too right?
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y=a/x is another matter. You HAVE to draw it on both sides. Btw this graph isn't in Emaths. It's amaths.
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paper 1 usually test what ah?
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really? but it's in my syllabus (4017) and there's still y=a/x^2, y=a^x and y=ax^3
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Originally posted by Garrick_3658:
y=a/x is another matter. You HAVE to draw it on both sides. Btw this graph isn't in Emaths. It's amaths.
Eh but I thought the syllabus state that you are supposed to know the graphs of y=ax^n
where n = -2,-1,1,2,3.
See http://www.seab.gov.sg/SEAB/oLevel/syllabus/2008_GCE_O_Level_Syllabuses/4016_2008.pdf
y= a/x ========> y = ax^-1
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to iampebble:
like if i say y = (x+1)(x-3), the coefficient of x^2 is definitely 1
erm you mean (x)*(x) = x^2 postive? or did you take (1)*(x)? so what you're trying to say that we shld look at these?
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ok mine's 4016 but if you really wanna know... i have some knowledge of it since i took amaths as well.
Simplest is to sub in ANY x value to get the corresponding y value, if comes out in exams to sketch a graph with an equation that you have never seen before, just do that.
y=a/x is a rectangular hyperbola
http://www.krysstal.com/images/graphs_hyperbola.gif
y=a/x^2 just flip all over to positive since (-x)^2 = +x^2 anyway
y=ax^3
http://www.krysstal.com/images/graphs_cubic.gif
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Originally posted by Mikethm:
Eh but I thought the syllabus state that you are supposed to know the graphs of y=ax^n
where n = -2,-1,1,2,3.
See http://www.seab.gov.sg/SEAB/oLevel/syllabus/2008_GCE_O_Level_Syllabuses/4016_2008.pdf
y= a/x ========> y = ax^-1
Ok bring it on :D
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urmm... thanks garrick but it's more confusing for me. so i guess ill stick to what iampebble/eagle explained.
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urmm anyone/?? no one answered my question :(
like if i say y = (x+1)(x-3), the coefficient of x^2 is definitely 1
erm you mean (x)*(x) = x^2 postive? or did you take (1)*(x)? so what you're trying to say that we shld look at these?
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Originally posted by gerlynn:
to iampebble:
like if i say y = (x+1)(x-3), the coefficient of x^2 is definitely 1
erm you mean (x)*(x) = x^2 postive? or did you take (1)*(x)? so what you're trying to say that we shld look at these?
Hello hello, I hope you're online, anyway!
Let's say we expand (x+1)(x-3):
x^2 - 3x + x - 3
= x^2 - 2x -3As I've mentioned, for a curve (which involves a x^2), you can see from what I expanded that the coefficient for x^2 is this case is 1, which is a positive number. That's why I mentioned that without expansion, you can see it is whether positive or negative because whether it is positive or negative depends on the coefficients of your x, because to get x^2, you use x*x, so if we have -x * x, it will be -x^2, which means coefficient is -1, which is negative. If we have -2x*x = -2x^2, which means coefficient is -2, which is negative.
Simply put, if they give you a equation:
y = 2x^2 + 3x + 1
You can deduce that it is a smiling graph, why? Because coefficient of x^2 is 2,which is positive.
If they give you something not so straightforward:
y = (x+2)(x-3)
Yes, it will be troublesome to expand it out, and moreover it's just sketching so it's even more unnecessary. But from that equation, how do you end up with
"ax^2 + bx + c"? Bx + c is not of your concern at all, because you're looking at a which is infront of your x^2, and the ONLY way you can get x^2 is only x * x, you can't get x^2 from x * 2, you will get 2x, right? So you only have to look at the COEFFICIENTS of the x's in your equation.y = (x+2)(x-3)
is also
y = (1x+2)(1x-3)
1x * 1x = 1x^2
O.o I hope I'm not confusing you, seriously..
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omg thank you so much!!!!!!!!!!! now i've clearly understand this! :D!!!!!!!!!
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Haha, miraculously you still understood it, which isn't a bad thing though, good luck for tomorrow! :D