Transformations of graphs!

**yoman360** · July 11th 2009, 01:49 PM

The graph of $y=sqrt(3x-x^2)$ is given. Use transformations to create a function whose graph is as shown.

1.)

**VonNemo19** · July 11th 2009, 02:29 PM

It looks like a reflection in the x axis (ie f(x)=-y) followed by vertical (1unit down) and horizontal (4 units back) shifts in the negative direction.

**yoman360** · July 11th 2009, 02:32 PM

Originally Posted by VonNemo19

It looks like a reflection in the x axis followed by vertical (1unit down) and horizontal 4 units back) shifts in the negative direction.

I'm trying to find the equation but I keep getting the wrong answer.

Here's the answer if it helps. I just don't know how to get that answer.
1) y= -sqrt(-x^2-5x-4)-1

Help!

**VonNemo19** · July 11th 2009, 02:41 PM

Originally Posted by yoman360

I'm trying to find the equation but I keep getting the wrong answer.

Here's the answer if it helps. I just don't know how to get that answer.
1) y= -sqrt(-x^2-5x-4)-1

Help!

If I had $y=\sqrt{x}$ . To flip it through the x axis I'd tack on a negative like so

$y=-\sqrt{x}$

If I wanted to slide it to the left 4 units I'd add four to the variable like this

$y=-\sqrt{(x+4)}$

If I then wanted it to shift down a unit I'd add a constant (subtract in this case)

$y=-\sqrt{(x+4)}-1$

Get it?

**yoman360** · July 11th 2009, 02:44 PM

Originally Posted by VonNemo19

If I had $y=\sqrt{x}$ . To flip it through the x axis I'd tack on a negative like so

$y=-\sqrt{x}$

If I wanted to slide it to the left 4 units I'd add four to the variable like this

$y=-\sqrt{(x+4)}$

If I then wanted it to shift down a unit I'd add a constant (subtract in this case)

$y=-\sqrt{(x+4)}-1$

Get it?

Ok I get it now I'm gonna try and see if I can get the answer. I'll put a post to tell you if i get it right or not.

**VonNemo19** · July 11th 2009, 02:51 PM

Originally Posted by yoman360

Ok I get it now I'm gonna try and see if I can get the answer. I'll put a post to tell you if i get it right or not.

Sounds like a plan.

**yoman360** · July 11th 2009, 02:52 PM

Ah I didn't get it right

this is what i did!
$y= sqrt(3x-x^2)$
$y= sqrt(-(3x-x^2))$
$y= -(sqrt(-(3x-x^2)))$
$y= -(sqrt(-(3x-x^2+1)))-1$
My answer: $y= - sqrt(-3x+x^2+1) - 1$

correct answer: $y=-sqrt(-x^2-5x-4)-1$

what did I do incorrectly?

**VonNemo19** · July 11th 2009, 02:59 PM

Originally Posted by yoman360

what did I do incorrectly?

Since you gave it the old college try I'll do it for you.

$y=-\sqrt{3(x+4)-(x+4)^2}-1$

**yoman360** · July 11th 2009, 03:06 PM

Originally Posted by VonNemo19

Since you gave it the old college try I'll do it for you.

$y=-\sqrt{3(x+4)-(x+4)^2}-1$

Thanks! But I don't understand how you got that.

**VonNemo19** · July 11th 2009, 03:09 PM

Originally Posted by yoman360

Thanks! But I don't understand how you got that.

To flip it, I tacked on the negative.

To slide to the left, i added 4 wherever i saw an x.

to slide it down, i subtracted 1 from everything.

**yoman360** · July 11th 2009, 03:12 PM

Originally Posted by VonNemo19

To flip it, I tacked on the negative.

To slide to the left, i added 4 wherever i saw an x.

to slide it down, i subtracted 1 from everything.

Oh thanks thats why! I relected y axis then shifted to the left 1 then reflected x axis then shift down 1.

**Plato** · July 11th 2009, 03:13 PM

Originally Posted by yoman360

Thanks! But I don't understand how you got that.

First 'move' it back 4 units by using $(x+4)$ for $x$ .

Flip it by multipling by $-1$ .

'Move' it down by subtracting $1$ .

**VonNemo19** · July 11th 2009, 03:14 PM

Originally Posted by yoman360

Oh thanks.

You're welcome

. You can click them buttons whenever you feel like it yoman.

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