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Maths AS question (Curves)

ImageUploadedByStudent Room1446937610.318693.jpg

I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance! :smile:


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This is not the answer, but your handwriting is amazing - do you use parker pens/expensive pens?
Reply 2
Avatar for Zacken
Zacken
22
Original post by KK.Violinist
ImageUploadedByStudent Room1446937610.318693.jpg

I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance! :smile:


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Your part (a) is slightly incorrect, you have −(x−c)(x−c)=−(x2−2cx+c2)-(x-c)(x-c) = -(x^2 - 2cx + c^2) (note the square of the cc) - now just distribute the −- sign into the brackets.
Work out all the letters in the equations then find B with making f(X) = g(X). C is the X intercept of h(X) so that should be easy and then length from that.
Reply 4
Avatar for Zacken
Zacken
22
Original post by KK.Violinist
ImageUploadedByStudent Room1446937610.318693.jpg

I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance! :smile:


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Note that f′(0)=f'(0) = gradient of line 1 - can you then write down the equation of line 1 (you know where it intersects the y-axis as well) and find where it intersects g(x)g(x)? What about finding where g(x)g(x) and f(x)f(x) intersect and then can you use g′(x)g'(x) to find the gradient of ℓ2\ell_2?
Original post by TheYearNiner
This is not the answer, but your handwriting is amazing - do you use parker pens/expensive pens?


Thanks haha, I use a standard ballpoint pen. :smile:


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Original post by Zacken
Note that f′(0)=f'(0) = gradient of line 1 - can you then write down the equation of line 1 (you know where it intersects the y-axis as well) and find where it intersects g(x)g(x)? What about finding where g(x)g(x) and f(x)f(x) intersect and then can you use g′(x)g'(x) to find the gradient of ℓ2\ell_2?


Sorry, my minds gone blank. How would I go about finding the equation of line 1? y=mx+c where -8 is c and mx being f'(x)?


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Avatar for kkboyk
kkboyk
21
Original post by KK.Violinist
Sorry, my minds gone blank. How would I go about finding the equation of line 1? y=mx+c where -8 is c and mx being f'(x)?


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Differentiate, sub in the x value of the coordinate of the point given which lies on line to get the gradient; and use y-y1 =m(x-x1).

Your handwriting is so damn good, it makes me feel like a toddler :eek:
Original post by kkboyk
Differentiate, sub in the x value of the coordinate of the point given which lies on line to get the gradient; and use y-y1 =m(x-x1).

Your handwriting is so damn good, it makes me feel like a toddler :eek:


Okay, if my calculations are correct then the line is y=2x-8.

Why is everyone talking about my handwriting!!! Haha I think it's pretty average, I've seen classmates with better handwriting xD


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Reply 9
Avatar for kkboyk
kkboyk
21
Original post by KK.Violinist
Okay, if my calculations are correct then the line is y=2x-8.

Why is everyone talking about my handwriting!!! Haha I think it's pretty average, I've seen classmates with better handwriting xD


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Because most people doing Maths (especially university Maths students and Professors) have terrible handwriting.

I sometimes can't read my own handwriting :frown:
Original post by kkboyk
Because most people doing Maths (especially university Maths students and Professors) have terrible handwriting.

I sometimes can't read my own handwriting :frown:


Lol that's true but surprisingly my teacher has very neat handwriting that makes even mine look mediocre.

Anyway, back on topic, now I have to find point A? I really shouldn't be doing this so late during the night. My brain keeps switching off -_-


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Reply 11
Avatar for kkboyk
kkboyk
21
Original post by KK.Violinist
Lol that's true but surprisingly my teacher has very neat handwriting that makes even mine look mediocre.

Anyway, back on topic, now I have to find point A? I really shouldn't be doing this so late during the night. My brain keeps switching off -_-


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Simultaneous equations with line L1 and g(x).
Original post by kkboyk
Simultaneous equations with line L1 and g(x).


I'm slightly confused by the D. I found D=40 but not sure if I'm correct. Do I even need to find D? Sorry for my amateur replies :frown:


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Reply 13
Avatar for kkboyk
kkboyk
21
Original post by KK.Violinist
I'm slightly confused by the D. I found D=40 but not sure if I'm correct. Do I even need to find D? Sorry for my amateur replies :frown:


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What is D?
Original post by kkboyk
What is D?


g(x)=x^2-14x+D




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Reply 15
Avatar for kkboyk
kkboyk
21
Original post by KK.Violinist


Oh its a constant.

If you differentiate g(x) and also L1 make them equal to each other (as they both will have the same gradient at that point) you'll be able to find the x value
Original post by kkboyk
Oh its a constant.

If you differentiate g(x) and also L1 make them equal to each other (as they both will have the same gradient at that point) you'll be able to find the x value


So find the second derivative for f(x) and the first for g(x) and equal to each other? If so then I got x=8


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