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#1
Hi guys,

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:
Attachment 738542738548

However why can't the graph be:
Attachment 738542738548738600
0
3 years ago
#2
(Original post by sienna2266)
Hi guys,

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:

However why can't the graph be:
Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.

Not like it matters anyway, it's all just for the purposes of working out the same region.
1
3 years ago
#3
(Original post by sienna2266)
Hi guys,

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:
Attachment 738542738548

However why can't the graph be:
Attachment 738542738548738600
both graphs are fine... in the top graph you are looking for values of x where the curve lies above ( or on ) the x axis

in the lower graph you are looking for values of x where the curve lies below ( or on ) the x axis
0
3 years ago
#4
(Original post by the bear)
both graphs are fine... in the top graph you are looking for values of x where the curve lies above ( or on ) the x axis

in the lower graph you are looking for values of x where the curve lies below ( or on ) the x axis
I would disagree, the equation given in the question is a negative quadratic and therefore the graph you sketch should portray this.
0
3 years ago
#5
(Original post by u12davisd)
I would disagree, the equation given in the question is a negative quadratic and therefore the graph you sketch should portray this.
it is not an equation
0
3 years ago
#6
(Original post by u12davisd)
I would disagree, the equation given in the question is a negative quadratic and therefore the graph you sketch should portray this.
But you don't have to draw the negative quadratic graph. The OP has clearly taken all terms on the other side of the inequality and correctly drawn the corresponding graph. The problem can be solved either way...
It doesn't specify you have to draw a specific quadratic graph- "use a sketch of appropriate quadratic graphs..."
1
3 years ago
#7
(Original post by sienna2266)
Hi guys,

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:
Attachment 738542738548

However why can't the graph be:
Attachment 738542738548738600
The roots remain the same because you just changed signs. However by changing signs you change the shape of the graph, you drew -f(x) rather than f(x). Hope that made sense. I tend to make that mistake too.
1
3 years ago
#8
(Original post by Anonymouspsych)
But you don't have to draw the negative quadratic graph. The OP has clearly taken all terms on the other side of the inequality and correctly drawn the corresponding graph. The problem can be solved either way...
It doesn't specify you have to draw a specific quadratic graph- "use a sketch of appropriate quadratic graphs..."
Fair enough, I would just recommend sketching the graph that is implied by the question as it is better practice for questions which involve having to get the sign correct or marks are lost. In this situation however this is not as important, obviously.
0
#9
(Original post by RDKGames)
Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.
Not like it matters anyway, it's all just for the purposes of working out the same region.
Ahh thanks so much for replying So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/sho...5154306&page=2
For this particular question(in the link above):
We or I came to the conclusion that the graph can be sketched in 2 ways:
First way:

Attachment 738556738558
However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

0
#10
Hey guys, thanks for your replies!
Here is a very similar question which is rather confusing:

So basically I understand they're asking you to "sketch" to just find out the solutions to the question but what I am specifically wondering about is the correct sketch of the graph. This one above is rather confusing
Please check my post above if possible- it would be much appreciated 0
3 years ago
#11
(Original post by sienna2266)
Hey guys, thanks for your replies!
Here is a very similar question which is rather confusing:

So basically I understand they're asking you to "sketch" to just find out the solutions to the question but what I am specifically wondering about is the correct sketch of the graph. This one above is rather confusing
Please check my post above if possible- it would be much appreciated If the coefficient of x in x^2 is positive it will form a u-shaped graph and if it's negative it will form an n-shaped graph.
0
#12
(Original post by dont know it)
If the coefficient of x in x^2 is positive it will form a u-shaped graph and if it's negative it will form an n-shaped graph.
Thanks but I am specifically talking about there being one graph on the axis not the one in the mark scheme. So this is not correct right:

It has to be the other way round? - so like n shaped?
1
3 years ago
#13
(Original post by sienna2266)
Thanks but I am specifically talking about there being one graph on the axis not the one in the mark scheme. So this is not correct right:

It has to be the other way round? - so like n shaped?
No that is correct. By rearranging you get x^2+x-2>0. Since the coefficient of x is positive, you get a U-shape, so that's right.
0
3 years ago
#14
(Original post by dont know it)
No that is correct. By rearranging you get x^2+x-2>0. Since the coefficient of x is positive, you get a U-shape, so that's right.
" x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0. "

Just copy and pasted what OP did. I don't think the graph is correct cos this is the graph -x^2 -x +2<0 . The negative in front x^2 means it's got a maximum as opposed to a minimum point. The one OP has drawn is x^2 +x -2>0 and so the graph has been reflected which is wrong I think.
But that's just my gut instinct -hopefully some of the others will confirm 1
3 years ago
#15
(Original post by h26)
" x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0. "

Just copy and pasted what OP did. I don't think the graph is correct cos this is the graph -x^2 -x +2<0 . The negative in front x^2 means it's got a maximum as opposed to a minimum point. The one OP has drawn is x^2 +x -2>0 and so the graph has been reflected which is wrong I think.
But that's just my gut instinct -hopefully some of the others will confirm You can do it either way I think depending on however you rearranged it. Also, if you multiply that equation by -1, you get x^2+x-2>0.
0
#16
(Original post by RDKGames)
Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.
Not like it matters anyway, it's all just for the purposes of working out the same region.
It would be extremely appreciated if you could please kindly reply back as you explain things so well.

Ahh thanks so much for replying So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/sho...5154306&page=2
For this particular question(in the link above):
We or I came to the conclusion that the graph can be sketched in 2 ways:
First way:

Second way:
Attachment 738574
However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

I just felt I needed to refer back to this one as it contradicts the question in this thread
0
3 years ago
#17
(Original post by sienna2266)
It would be extremely appreciated if you could please kindly reply back as you explain things so well.

Ahh thanks so much for replying So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/sho...5154306&page=2
For this particular question(in the link above):
We or I came to the conclusion that the graph can be sketched in 2 ways:
First way:
However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

I just felt I needed to refer back to this one as it contradicts the question in this thread
It's not *wrong* it's just different graphs representing the same solutions to the inequality. If you want to sketch what the question lays out then yes you need to strictly follow the first way, but generally it doesn't matter.
0
#18
(Original post by RDKGames)
It's not *wrong* it's just different graphs representing the same solutions to the inequality. If you want to sketch what the question lays out then yes you need to strictly follow the first way, but generally it doesn't matter.
Thanks so much for replying. But if we ignore working out the solutions to the inequality and focus on drawing the correct sketch of the graph - doesn't that mean that the graph for way 2(alternative graph i thought of) should have a upside down U shaped graph? Btw I understand the graph shown in the mark scheme is correct. I am just wondering if the the alternative graph should be U shaped or upside down U shaped?
This is what I mean by upside down U shaped: and this is caused by -x^2 -x +2<0 -the solutions would be x<-2 or x>1

This is what I mean by U shaped graph: and this is caused by x^2 +x -2>0 - the solutions would be be x<-2 or x>1
Attachment 738576738578
Both these graphs are from x^2 +8<2x^2 +x +6 and these graphs are not from the mark scheme -just an alternative graph i think could work
So I understand both these graphs yield the same solutions and they are both correct to use to work out the solutions. But what I am specifically asking is which one would be the correct representation/sketch of the graph x^2 +8<2x^2 +x +6?

Would be really really appreciated if you could please reply back. Thank you so much!
0
3 years ago
#19
(Original post by sienna2266)
But what I am specifically asking is which one would be the correct representation/sketch of the graph x^2 +8<2x^2 +x +6?

Would be really really appreciated if you could please reply back. Thank you so much!
Neither or both, whichever way you want to think about it. But the point is that one does not take precedence over the other, they are both equally as fine in order to answer the original inequality and both are fine to sketch in order to answer it.
0
#20
(Original post by RDKGames)
Neither or both, whichever way you want to think about it. But the point is that one does not take precedence over the other, they are both equally as fine in order to answer the original inequality and both are fine to sketch in order to answer it.
But why not for the question here then: I've quoted what you said about it below

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:

However why can't the graph be:
Attachment 738584738586

(Original post by RDKGames)
Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.

Not like it matters anyway, it's all just for the purposes of working out the same region.
0
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