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September 21, 2025, 09:44:27 am

Author Topic: f'ns and transformations  (Read 890 times)  Share 

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nacho

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f'ns and transformations
« on: February 08, 2011, 09:04:19 pm »
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Hi,
These are out of the year 12 maths quest book, pages 64 - 65:

11. For the parabola whos range is y <= 3, whose x-coordinate of the turning point is -4 and whose y-intercept is y = -2 1/3, find:

a. the y-coordinate of the turning point
b. the equation of the parabola
c. the coordinates of the x-intercepts
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kaushik

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Re: f'ns and transformations
« Reply #1 on: February 08, 2011, 09:34:17 pm »
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y = a(x − h)2 + k
a) Range is y ≤ 3 ⇒ a < 0
   and k = 3 is the y-coordinate of
   turning point.
   h = −4

edit i can't do the others without latex. how do i use latex?  :P

xZero

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Re: f'ns and transformations
« Reply #2 on: February 08, 2011, 09:35:39 pm »
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a.the given range should tell you that the parabola is reflected in the x-axis and the highest value within the range should be the y-coordinate of the turning point y=3

b.the general form of a parabola is
then differentiate the equation to get , since the turning point is at , we can sub it into the derivative to get 

sub into so you get , then sub and into the equation to find a, then you can find b from there

c. should be straight forward

this was under the assumption that the parabola was inverted, someone verify my working please
« Last Edit: February 09, 2011, 02:28:50 pm by xZero »
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xZero

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Re: f'ns and transformations
« Reply #3 on: February 08, 2011, 09:43:02 pm »
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y = a(x − h)2 + k
a) Range is y ≤ 3 ⇒ a < 0
   and k = 3 is the y-coordinate of
   turning point.
   h = −4

edit i can't do the others without latex. how do i use latex?  :P

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nacho

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Re: f'ns and transformations
« Reply #4 on: February 08, 2011, 10:25:27 pm »
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b.the general form of a parabola is
then differentiate the equation to get , since the turning point is at , we can sub it into the derivative to get  

sub into so you get , then sub and into the equation to find a, then you can find b from there
thanks!
T_T can't believe i didn't realise question a...
Anyway, for b, is there another way? ALthough it's assumed knowledge, i feel MQuest isn't the type to ask a question in which calculus is involved, especially in ch. 2
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xZero

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Re: f'ns and transformations
« Reply #5 on: February 08, 2011, 10:33:19 pm »
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b.the general form of a parabola is
then differentiate the equation to get , since the turning point is at , we can sub it into the derivative to get  

sub into so you get , then sub and into the equation to find a, then you can find b from there
thanks!
T_T can't believe i didn't realise question a...
Anyway, for b, is there another way? ALthough it's assumed knowledge, i feel MQuest isn't the type to ask a question in which calculus is involved, especially in ch. 2

well i did part b before i realised part a LOL but anyways use (T.P form), sub in to find a and viola, part B's done

Edit: apparently if you torrenting while posting you can triple post without knowing  ???
« Last Edit: February 09, 2011, 02:28:06 pm by xZero »
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Re: f'ns and transformations
« Reply #6 on: February 09, 2011, 09:04:54 am »
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*voila :P
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