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fredrick · « **on:** October 18, 2008, 08:15:29 pm »

The graph of $y=2(x+3)(x+1)(x-1)$ is reflected in the y-axis, then in the x axis and translated 2 units in the positive x direction to give h(x).
Show that h(x)=2(x-5)(x-3)(x-1)

Collin Li · « **Reply #1 on:** October 18, 2008, 08:18:48 pm »

Reflecting in the y-axis requires that all $x$ are replaced with $-x$ :

$2(3-x)(1-x)(-x-1) = -2(x-3)(x-1)(x+1)$

Reflecting in the x-axis requires that the entire function is multiplied by $-1$ :

$2(x-3)(x-1)(x+1)$

Translating 2 units in the positive x-direction requires that all instances of $x$ are replaced with $x-2$ :

Hence: $h(x) = 2(x-2 - 3)(x-2 - 1)(x-2 + 1) = 2(x-5)(x-3)(x-1)$

bucket · « **Reply #2 on:** October 18, 2008, 08:21:32 pm »

If there was a translation along the y axis, say 5 in the positive, would you just add +5 to the end of the whole equation or is there a different way to do that with these forms?

Collin Li · « **Reply #3 on:** October 18, 2008, 08:23:06 pm »

Yeah, you add 5 to the whole equation.

Mao · « **Reply #4 on:** October 18, 2008, 08:29:32 pm »

Quote from: bucket on October 18, 2008, 08:21:32 pm

If there was a translation along the y axis, say 5 in the positive, would you just add +5 to the end of the whole equation or is there a different way to do that with these forms?

as a general rule:

when dilating parallel to an axis by a factor of a, replace the axis variable by that variable divided by a
i.e. dilating parallel to x by a factor of 2: $x\to \frac{x}{2}$
dilating parallel to y by a factor of 2: $y\to \frac{y}{2}$

when translating in a certain direction by c units, replace the axis variable by that variable minus c
i.e. translating in the x direction two units to the left: $x \to x-(-2)$
translating in the y direction five units up: $y \to y-5$

when reflecting in an axis, replace the other axis variable by its negative
i.e. reflecting in y: $x\to -x$
reflecting in x: $y\to -y$

note that for the above cases of y, rearranging will give the forms you are familiar with $y=(-)af(x)+c$ , experiment with these, you'll find that this is a LOT easier to remember.

[strictly speaking, this is also more "correct". transformations are best described by transformation matrices (CAS students slightly touch on that), and the answer is then rearranged to express y explicitly in terms of x.]

Toothpaste · « **Reply #5 on:** October 18, 2008, 08:32:09 pm »

Quote from: Toothpick on March 15, 2008, 09:52:17 pm

Quote from: bec on March 15, 2008, 09:30:12 pm
this is kind of a strange question, but why does f(x-h) translate the graph of f(x) h units to the RIGHT?
i know that is does , and i know how to apply it but it's just something i've rote-learned, i don't actually understand why.

take the graph of f(x)=x+2 for example:
to translate it one unit to the right, i know we need to make f(x-1) = (x-1)+2 = x+1
but when a graph is "moved" 1 unit to the right, aren't we, in effect, adding 1 to every x-coordinate? i know that's completely wrong, but logically it seems right.

did that make any sense?

For example, let's say we have $y = x^2$ and we want to shift it in the positive direction of the x-axis
$(x,y) \rightarrow (x+1,y)$ (Here we add 1 to x, so yes, we are adding 1 to every x-coordinate.)
i.e. original $\rightarrow$ new
let $x_1$ and $y_1$ be the new points/new image.

$x_1 = x+1$ and $y_1 = y$
$x= x_1 - 1$ and $y = y_1$

Substitute x and y into your equation.

$y_1 = (x_1 -1)^2$
$\therefore \implies y=(x-1)^2$ is the new image.

Reflections about the x axis.
I'll use $y = x^2$ again.
$(x,y) \rightarrow (x,-y)$ (why? see next example for an explanation)
let $x_1$ and $y_1$ be the new points/new image.
$x_1 = x$ and $y_1 = -y$

$x = x_1$ and $y = -y_1$

Our equation: $y = x^2$
$-y_1= x_1^2$
$y_1 = -x_1^2$
$\therefore y = -x^2$

Reflections about the y axis.
$y = \sqrt{x}$
$(x,y) \rightarrow (-x,y)$

(It's -x because we're changing the x values when we reflect, not the y. Visualise the graph. It's sticking out to the positive side of the x axis right? When we reflect it in the y-axis all the x-values are on the negative side now, and the y-values remain the same.)

let $x_1$ and $y_1$ be the new points/new image.
$x_1=-x$ and $y_1 = y$

$x= -x_1$

$y = \sqrt{x}$
$y_1=\sqrt{-x_1}$
$\therefore y = \sqrt{-x}$

Dilation of factor 5 from the x-axis.
$y = \frac{1}{x^2}$
$(x,y) \rightarrow (x,5y)$

(Same reason why it's 5y. We're, in general, 'pulling' the graph up/down - changing the y values and not the x)

let $x_1$ and $y_1$ be the new points/new image.
$x_1 = x$ and $y_1 = 5y$

$\frac{y_1}{5} = y$

$\frac{y_1}{5} = \frac{1}{(x_1)^2}$
$y_1 = \frac{5}{(x_1)^2}$
$\therefore y = \frac{5}{(x)^2}$ is the new image.

Might help

Mao · « **Reply #6 on:** October 18, 2008, 08:35:12 pm »

Quote from: Toothpick on March 15, 2008, 09:52:17 pm

Dilation of factor 5 from the x-axis.
$y = \frac{1}{x^2}$
$(x,y) \rightarrow (x,5y)$

(Same reason why it's 5y. We're, in general, 'pulling' the graph up/down - changing the y values and not the x)

let $x_1$ and $y_1$ be the new points/new image.
$x_1 = x$ and $y_1 = 5y$

$\frac{y_1}{5} = y$

$\frac{y_1}{5} = \frac{1}{(x_1)^2}$
$y_1 = \frac{5}{(x_1)^2}$
$\therefore y = \frac{5}{(x)^2}$ is the new image.

wow, i like.

fredrick · « **Reply #7 on:** October 18, 2008, 08:41:22 pm »

I do it the mapping way but it didnt work out.

$(x,y)\to (-x+2,-y)$
$\acute{x}=-x+2 and \acute{y}=-y$
$x=2-\acute{x}$ and $y=-\acute{y}$

Now i sub them into the equation:
$h(x)=2(2-x+3)(2-x+1)(2-x-1)$
$h(x)=2(-x+5)(-x+3)(-x+1)$
so $h(x)=-2(x-5)(x-3)(x-1)$ Where di i go wrong

bucket · « **Reply #8 on:** October 18, 2008, 08:44:38 pm »

wow toothpick, thats good

fredrick · « **Reply #9 on:** October 18, 2008, 08:47:42 pm »

ohhhhhhh nvm i kno wat i did wrong! Im suppose to end up with:
$-h(x)=-2(x-5)(x-3)(x-1)$
Just forgot the negative sign :idiot2:

Edit: one more question say we have $-f(-\frac{1}{2}x+2)$
That means f(x) is reflected in both axis, dilated factor 2 from y axis and translated 4 units to the right, right?

bucket · « **Reply #10 on:** October 18, 2008, 08:52:46 pm »

I have a question, does anybody else get confused when they word the transformations differently??
ie. dilation to the x...dilation from the x... etc.
apparently they mean the opposite or something? It always throws me off and I don't know what type of dilation to do.

shinny · « **Reply #11 on:** October 18, 2008, 08:56:20 pm »

I've had some awful phrasing in transformations before. e.g. Insight 2006: 'dilation by a scale factor of 1/2 along the y-axis'...I interpreted that as $k(x)=\frac{1}{2}h(x)$ , but they wanted $k(x)=h(\frac{x}{2})$ o_O wth

bucket · « **Reply #12 on:** October 18, 2008, 08:57:45 pm »

YEAH I DID THAT ONE TOO!! thats exactly what im talking about!!

Mao · « **Reply #13 on:** October 18, 2008, 09:04:53 pm »

Quote from: shinjitsuzx on October 18, 2008, 08:56:20 pm

I've had some awful phrasing in transformations before. e.g. Insight 2006: 'dilation by a scale factor of 1/2 along the y-axis'...I interpreted that as $k(x)=\frac{1}{2}h(x)$ , but they wanted $k(x)=h(\frac{x}{2})$ o_O wth

that would be my interpretation also.

fredrick · « **Reply #14 on:** October 18, 2008, 09:05:35 pm »

?! i think they mite be wrong on that, even if it meant scale factor of 1/2 from the y -axis wouldnt that be $k(x)=h(2x)$ ?

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