E Maths: Graph + Cube Root

gerlynn SG

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?

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!

eagle SG

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)

Complete the square (you should be able to do)
If cannot, get crucifixed or pillow to practice doing and post the solution

Sketching 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 all

c) cube root [(0.57147 / 36783)]

Are you taking the old syllabus? New syllabus paper 1 can use calculator

davidche SG

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 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.

Eagle has got it.

c) cube root [(0.57147 / 36783)]

press the dam calculator.

gerlynn SG

fyi davidche, mine is the old syllabus.

there's a difference btw sketch and draw?

iamapebble SG

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.

jlzk SG

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

gerlynn SG

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?

Oceane SG

Deleted. Wrong equations.

iamapebble SG

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. =)

Oceane SG

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.

gerlynn SG

hi! sorry but i still dont get it :(

about the coefficient part and how to know which is which.

iamapebble SG

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

gerlynn SG

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?

Garrick_3658 SG

y=a/x is another matter. You HAVE to draw it on both sides. Btw this graph isn't in Emaths. It's amaths.

Maith SG

paper 1 usually test what ah?

gerlynn SG

really? but it's in my syllabus (4017) and there's still y=a/x^2, y=a^x and y=ax^3

Mikethm SG

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

gerlynn SG

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?

Garrick_3658 SG

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

Garrick_3658 SG

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