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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on March 14, 2009, 11:35:52 am

Title: Question that pisses e off
Post by: khalil on March 14, 2009, 11:35:52 am: man.. i've been trying to question for 3 hrs..still didn't get the answer....

-ax^4
the graph has been dilated by a factor of 2 parallel to the y axis and then translated h units to the left
the graph is now translated 1 unit upwards and as a result cuts the y axis at y=-3
if the curve passes through the point (-2,-3)
find the equation of the curve
Title: Re: Question that pisses e off
Post by: Flaming_Arrow on March 14, 2009, 11:41:47 am: $y= -ax^4$

$y= \frac{-a(x+h)^4}{16}+1$

our points are (0,-3) and (-2,-3)

$-3= \frac{-ah^4}{16}+1$

$-3= \frac{-a(-2+h)^4}{16}+1$

yeilds

$a=64 \ h=1$

$\therefore y= -4(x+1)^4}+1$ or like TrueTears answer $y = -4x^4 -16x^3 - 24x^2 - 16x -3$
Title: Re: Question that pisses e off
Post by: TrueTears on March 14, 2009, 11:46:35 am: (x,y) -> (x,2y) -> (x-h,2y) -> (x-h,2y+1)

so x = x'-h

and y = 2y'+1

solve for x' and y'

x' = x+h and $y'= \frac{1}{2}(y-1)$

sub these in $y = -ax^4$

$-a(x+h)^4 = \frac{1}{2}(y-1)$

now create 2 simultaneous equations using the info given ie the graph passes through (0,-3) and (-2,-3)

sub these in yields

$-a(0+h)^4 = \frac{1}{2}(-3-1)$ ...[1]

$-a(-2+h)^4 = \frac{1}{2}(-3-1)$ ...[2]

solve for a and h yields a = 2 and h = 1

so equation is $-2(x+1)^4 = \frac{1}{2}(y-1)$

you can solve for y which yields $y = -4x^4 -16x^3 - 24x^2 - 16x -3$
Title: Re: Question that pisses e off
Post by: Flaming_Arrow on March 14, 2009, 11:52:16 am: Quote from: TrueTears on March 14, 2009, 11:46:35 am
(x,y) -> (x,2y) -> (x-h,2y) -> (x-h,2y+1)

so x = x'-h

and y = 2y'+1

solve for x' and y'

x' = x+h and $y'= \frac{1}{2}(y-1)$

sub these in $y = -ax^4$

$-a(x+h)^4 = \frac{1}{2}(y-1)$

now create 2 simultaneous equations using the info given ie the graph passes through (0,-3) and (-2,-3)

sub these in yields

$-a(0+h)^4 = \frac{1}{2}(-3-1)$ ...[1]

$-a(-2+h)^4 = \frac{1}{2}(-3-1)$ ...[2]

solve for a and h yields a = 2 and h = 1

so equation is $-2(x+1)^4 = \frac{1}{2}(y-1)$

you can solve for y which yields $y = -4x^4 -16x^3 - 24x^2 - 16x -2$

we both answered the same question xD
Title: Re: Question that pisses e off
Post by: TrueTears on March 14, 2009, 11:53:45 am: lol yeah haha
Title: Re: Question that pisses e off
Post by: khalil on March 14, 2009, 06:47:38 pm: Thanks guys but why cant i do i get an unknown result when i simplify the equations?
using -ah^4=-2 and -a(h-2)^4=-2