Algebra - transformations of quadratic functions

I don't understand how y = 4x (to the power of 2) -1 is a dilation from the y axis by 1/2. but y = 2x(to the power of 2) is a dilation from the x-axis by 2??????

I also need help with the transformation of y = x(power 2) to y = (2x - 2) power 2 + 3

Thanks

Re: Algebra - transformations of quadratic functions

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Originally Posted by

**elmidge** I don't understand how y = 4x (to the power of 2) -1 is a dilation from the y axis by 1/2. but y = 2x(to the power of 2) is a dilation from the x-axis by 2??????

Looking at the base case of $\displaystyle y= x^2$, if x= 0, y= 0 and if x=-1, y= 1. For $\displaystyle y= 4x^2$ if x=0, y= 0 and if x= 1, y= 4. That is NOT a "dilation of the y-axis by 1/2" it is a dilation by 4, just as for [tex]y= 2x^2[/itex] it is a dilation of the **y** axis by 2. On the other hand, as x goes from 0 to 1/2, y goes from 0 to $\displaystyle 4(1/2)^2= 1$ so the first is a dilation of the **x** axis by 1/2. I think you have the x and y axes reversed!

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I also need help with the transformation of y = x(power 2) to y = (2x - 2) power 2 + 3

The "base" $\displaystyle y= x^2$ has vertex at x= 0, y= 0 and when x= 1 or -1, y= 1. $\displaystyle y= (2x-2)^2+ 3$ has vertex at 2x-2= 0 or x= 1 and y= 3. So, immediately, there is a translation from (0, 0) to (1, 3) When x= 2, (shifting to the right 1 just as from 0 to 1 before) 2x- 2= 2 so $\displaystyle y= 2^2+ 3= 7$. That is "y= 0 to y= 1", a difference of 1, has changed to "y= 3 to y= 7" a difference of 4. There is a dilation of the y-axis by 4. One can argue that this is because the "2x" squared has become $\displaystyle 4x^2$.