# Core 3 - Transformations questionWatch

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Thread starter 6 years ago
#1
Anyone care to explain the solution to this question, transformations is my weakest area and I'm trying to get my head around it... But this has stumped me :P Thanks.

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6 years ago
#2
(Original post by ashboot)
Anyone care to explain the solution to this question, transformations is my weakest area and I'm trying to get my head around it... But this has stumped me :P Thanks.

I'm not sure, but by completing the square I think it is 2 units to the left and 4 units down. I don't think it is possible to say what the value of k is from the information shown.

Where is the question from?
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6 years ago
#3
(Original post by ashboot)
...
I'd start by completing the square inside the brackets, and then see how you can use those three transformations.
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6 years ago
#4
(Original post by steve2005)
I don't think it is possible to say what the value of k is from the information shown.
You can't determine k until part iii
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6 years ago
#5
(Original post by ghostwalker)
You can't determine k until part iii
OK Thanks.

So the stretch is times k parallel to the y-axis and the range is greater than or equal to -4k.
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6 years ago
#6
(Original post by steve2005)
OK Thanks.

So the stretch is times k parallel to the y-axis and the range is greater than or equal to -4k.
Agreed.
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6 years ago
#7
(Original post by ghostwalker)
Agreed.

I get k = 5 and x= -2(root)2 -2, -2 and 2(root2)-2.

I think this is a nasty question but it is interesting.

EDIT

It looks as if my answer is wrong.

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6 years ago
#8
(Original post by steve2005)
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I agree, and thought the question was really neat. Gets one to think rather than just rote bookwork.
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6 years ago
#9
(Original post by ghostwalker)
I agree, and thought the question was really neat. Gets one to think rather than just rote bookwork.
Agreed. I remember that 99% of questions I did at A-level (inc FM) were plain, so you have to be really lucky to get a neat question.
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6 years ago
#10
I have found my mistake in the graph.

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6 years ago
#11
(Original post by ashboot)
Anyone care to explain the solution to this question, transformations is my weakest area and I'm trying to get my head around it... But this has stumped me :P Thanks.

9i) yes begin with completing the square, to get k[(x+2)^2-4]
So from y=x^2 to y=(x+2)^2 is a translation parallel to the x axis since the +2 is inside the function.
From y=x^2 to y=(x+2)^2 -4 is a translation parallel to y axis as -4 is outside the function.
From y=x^2 to the k[(x+2)^2-4] is then a stretch of k. Hope this helps!!

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6 years ago
#12
9ii) since completing the square you should know that the graph hits y=-4k, since expending this k[(x+2)^2-4] gives k(x+2)^2-4k and so the range must be y is GREATER THAN OR EQUAL TO -4k.

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6 years ago
#13
Nice question!

I got
i) 2 left
4 down
Stretch of scale factor k parallel to the y axis

ii) f(x) = k(x+2)^2 - 4k so minimum is -4k when x = -2. So f must be greater than or equal to -4k.

iii) Sketch graph of |f(x)|, if there are 3 values for |f(x)| = 20 then it must touch the top of the maxima, which is now 4k. 4k = 20 so k = 5.
5x^2 + 20x = 20
5x^2 + 20x - 20 = 0
x^2 + 4x - 4 = 0

x = 2 + 2root2
x = 2 - 2root2
x = - 2 (from part ii)
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6 years ago
#14
Well, I hope the OP appreciated all that.

Slap wrists guys - some spoilerification (is that a word?) would not go amiss.
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