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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: nacho on February 08, 2011, 09:04:19 pm

Title: f'ns and transformations
Post by: nacho on February 08, 2011, 09:04:19 pm: 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
Title: Re: f'ns and transformations
Post by: kaushik on February 08, 2011, 09:34:17 pm: 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
Title: Re: f'ns and transformations
Post by: xZero on February 08, 2011, 09:35:39 pm: 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 $ax^2+bx+c, when\ x=0, y=-2\frac{1}{3}, so\ c= -2\frac{1}{3}$
then differentiate the equation to get $2ax+b$ , since the turning point is at $x=-4$ , we can sub it into the derivative to get $-8a+b=0$

sub $b=8a$ into $ax^2+bx-2\frac{1}{3}$ so you get $ax^2+8ax-2\frac{1}{3}$ , then sub $x=-4$ and $y=3$ 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
Title: Re: f'ns and transformations
Post by: xZero on February 08, 2011, 09:43:02 pm: Quote from: kaushik on February 08, 2011, 09:34:17 pm
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

http://vce.atarnotes.com/forum/index.php/topic,3137.0.html
Title: Re: f'ns and transformations
Post by: nacho on February 08, 2011, 10:25:27 pm: Quote from: xZero on February 08, 2011, 09:35:39 pm
b.the general form of a parabola is $ax^2+bx+c, when\ x=0, y=-2\frac{1}{3}, so\ c= -2\frac{1}{3}$
then differentiate the equation to get $2ax+b$ , since the turning point is at $x=-4$ , we can sub it into the derivative to get $-8a+b=0$

sub $b=8a$ into $ax^2+bx+2\frac{1}{3}$ so you get $ax^2+8ax+2\frac{1}{3}$ , then sub $x=-4$ and $y=3$ 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
Title: Re: f'ns and transformations
Post by: xZero on February 08, 2011, 10:33:19 pm: Quote from: nacho on February 08, 2011, 10:25:27 pm
Quote from: xZero on February 08, 2011, 09:35:39 pm
b.the general form of a parabola is $ax^2+bx+c, when\ x=0, y=-2\frac{1}{3}, so\ c= -2\frac{1}{3}$
then differentiate the equation to get $2ax+b$ , since the turning point is at $x=-4$ , we can sub it into the derivative to get $-8a+b=0$

sub $b=8a$ into $ax^2+bx-2\frac{1}{3}$ so you get $ax^2+8ax-2\frac{1}{3}$ , then sub $x=-4$ and $y=3$ 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 $a(x+4)^2+3$ (T.P form), sub $[0,-2\frac{1}{3}]$ in to find a and viola, part B's done

Edit: apparently if you torrenting while posting you can triple post without knowing ???
Title: Re: f'ns and transformations
Post by: pHysiX on February 09, 2011, 09:04:54 am: *voila :P