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Sketching Curves - C1 - Chapter 4

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Original post by TeeEm
there should be no x, x was substituted with the x from each of the 2 coordinates from A and B


Hey, sorry, I've found another question that I need help on :P.

Coincidentally, it required the same formula and I've correctly figured out the answer to part a, partially.

ImageUploadedByStudent Room1450720604.768458.jpg

I got the answer of the equation of the line to be x^2 - 4x + 3 = 0.
I got this correct, but I assumed that this is an x^2 graph.
What if it was an 2x^2 graph, then it'd have been a different answer, how can I distinguish or find out what type of graph it is. Do I just assume that this is always an x^2 graph?

I could use what @ubisoft said before, (x-A)(x-B) and it would give me the exact same answer and confirm that it is an x^2 graph and not 2x^2 etc.

The answer to this question part a also said f greater than or equal to -1. Is this because of the minimum point?


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(edited 8 years ago)
Original post by Chittesh14


What if it was an 2x^2 graph, then it'd have been a different answer, how can I distinguish or find out what type of graph it is. Do I just assume that this is always an x^2 graph?


No. You know that it is of the form ax2+bx+cax^2+bx+c. You have three points. You need to use all of them.
Original post by BuryMathsTutor
No. You know that it is of the form ax2+bx+cax^2+bx+c. You have three points. You need to use all of them.


Oh okay :tongue:.

Hey, I came across here weird questions lol.

The answers are different too.

ImageUploadedByStudent Room1450722818.030224.jpg

After 2(f), so (g) and onwards, I got everything wrong.



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Original post by Chittesh14
Oh okay :tongue:.

Hey, I came across here weird questions lol.

The answers are different too.

ImageUploadedByStudent Room1450722818.030224.jpg

After 2(f), so (g) and onwards, I got everything wrong.



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Take 4x−5y=104x-5y=10 for example. If (0,c) is on the line you can find c by substituting 0 for x to get -5y=10 leading to y=-2.
(edited 8 years ago)
Original post by BuryMathsTutor
Take 4x−5y=104x-5y=10 for example. If (0,c) is on the line you can find c by substituting 0 for x to get -5y=10 leading to y=-2.

I sympathise with stuck students. Currently trying to find lower bounds on Hausdorff measures. :frown:


I'm not always stuck, but I do get stuck often and it's fun I just learn more tbh. You are really kinda you help a lot :smile:.

I know that way because even if I rearrange it:

5y = 4x-10
Y = 4/5x - 2

The y-intercept is -2.

But, just solve the y-intercepts for my question. Please, just do (g), (h) and (i).

I got the answers:

(g) = -4/7
(h) = 2/9
and (i) = -2/3

But, according to the answers, all these are wrong.


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Original post by Chittesh14
I'm not always stuck, but I do get stuck often and it's fun I just learn more tbh. You are really kinda you help a lot :smile:.

I know that way because even if I rearrange it:

5y = 4x-10
Y = 4/5x - 2

The y-intercept is -2.

But, just solve the y-intercepts for my question. Please, just do (g), (h) and (i).

I got the answers:

(g) = -4/7
(h) = 2/9
and (i) = -2/3

But, according to the answers, all these are wrong.


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g) 3x-4y+8=0
If x=0 then -4y+8=0 and so y=2

h) I already did this one..

i) -2x+y-9=0
If x=0 then y-9=0 and so y=9

Now try the others. Post your working if you like.
Original post by BuryMathsTutor
g) 3x-4y+8=0
If x=0 then -4y+8=0 and so y=2

h) I already did this one..

i) -2x+y-9=0
If x=0 then y-9=0 and so y=9

Now try the others. Post your working if you like.


Oops never mind, I didn't realise you already done that one.
Now I realise, that it's not always the y-intercept in the equation of the line. Sometimes, you have to solve for it, it's not always in the for. Of y = mx + c where c is the y-intercept. I kept taking c as the y-intercept in the equation of the line, sorry.


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Original post by BuryMathsTutor
g) 3x-4y+8=0
If x=0 then -4y+8=0 and so y=2

h) I already did this one..

i) -2x+y-9=0
If x=0 then y-9=0 and so y=9

Now try the others. Post your working if you like.


Sorry, big misunderstanding, I copied all the questions from 3(g) and onwards lol...


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Original post by BuryMathsTutor
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Original post by SeanFM
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Original post by TeeEm
...


ImageUploadedByStudent Room1450803719.268401.jpg

I need help on part a please.

My working out so far:

ImageUploadedByStudent Room1450803763.303154.jpg


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(edited 8 years ago)
Reply 29
Avatar for TeeEm
TeeEm
19
Original post by Chittesh14
ImageUploadedByStudent Room1450803719.268401.jpg

I need help on part a please.

My working out so far:

ImageUploadedByStudent Room1450803763.303154.jpg


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I am teaching until 22.00
Somebody else I am sure will help you until then.
Original post by TeeEm
I am teaching until 22.00
Somebody else I am sure will help you until then.


Ok, thank you and sorry for disturbing you :biggrin:.


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Original post by Chittesh14
ImageUploadedByStudent Room1450803719.268401.jpg

I need help on part a please.

My working out so far:

ImageUploadedByStudent Room1450803763.303154.jpg


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You're OK up to where you found the gradient. You then seem to be using the wrong point.

Also at this point it's probably easiest to use y−y1=m(x−x1)y-y_1=m(x-x_1) since you have the point and the gradient.
Original post by BuryMathsTutor
You're OK up to where you found the gradient. You then seem to be using the wrong point.

Also at this point it's probably easiest to use y−y1=m(x−x1)y-y_1=m(x-x_1) since you have the point and the gradient.


Sorry, I forgot to edit this post. I got the question correct :smile:, I've never come across this type of question xd similar ones, but not this one so I didn't know what to do. I substituted the values from B into that equation ^ and solved it about 20 mins after the post.


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Original post by BuryMathsTutor
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ImageUploadedByStudent Room1450812909.599442.jpg

Need help on part b please. I've done part a and I've got it correct.

ImageUploadedByStudent Room1450812988.978007.jpg

I am 99.9% sure that there is an error in the gradient, so I either shouldn't have assumed that the gradient is 1/2... In that working out

Because I've tried all the ways and I just get 11 when the gradient is 1/2.

EDIT: I've solved it, the right angled triangle - sides perpendicular, I'm such an idiot... Still learning!
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(edited 8 years ago)
Original post by BuryMathsTutor
....


Can you help me please? ^^


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Original post by Chittesh14
Can you help me please? ^^


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Help with what? Your edit above says that you have solved it.
Original post by BuryMathsTutor
Help with what? Your edit above says that you have solved it.


Oh wow, tagged you in the wrong thing -_-..


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Original post by BuryMathsTutor
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Original post by SeanFM
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Original post by TeeEm
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Original post by ubisoft
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Question: Find the value of n for which Un has the given value:

Attachment not found
(edited 8 years ago)
ImageUploadedByStudent Room1450891457.816439.jpg19.4KB
Original post by Chittesh14
Question: Find the U1, U2, U3 and U10 of the following sequences, where:
ImageUploadedByStudent Room1450891457.816439.jpg


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What have you tried so far? :smile:
Original post by SeanFM
What have you tried so far? :smile:


Sorry, wrong question ffs, I'm an idiot.

Question: Find the value of n for which Un has the given value:

Attachment not found


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(edited 8 years ago)

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