# Math Help - Transformations

1. ## Transformations

The graph of the function is given in blue. Use transformations to create the function whose graph is shown in red.

to

I don't know how to do it.
I tried

this hw is very important. please help me. thank you!

2. Originally Posted by greenbee
The graph of the function is given in blue. Use transformations to create the function whose graph is shown in red.

to

I don't know how to do it.
I tried

this hw is very important. please help me. thank you!
There are some handy tricks to do this, I'll help start you off here, and I strongly recommending reading Function Transformations / Translations: Basic Rules .

We know right off the bat that we have a vertical reflection across the x-axis. What does this do to our y-values? It reverses them! Anything that once was positive is now negative, and vice versa. So let's

Replace $y$ with $-y$

Next, we have a vertical stretch. Look at the blue vertex; $(1,1)$. Look at the Red one, (-2,0.5). Our vertex had its y-value halved! (Keep in mind, we know it is a stretch, and not a translation because the roots [invariant points on the x-axis] have not changed.) So lets

Replace $y$ with $\frac{1}{2}y$

Lastly, what else has this graph done? It moved left 3 units from $x=1$ to $x=-2$! Let's factor that in;

Replace $x$ with $(x+3)$.

Do your replacements and simplify it all back down to the form $y=f(x)$, and plug it into your graphing calculator to make sure!

3. i don't get it.

I got

y = -(sqrtX+6+(X+3)^2) / (1/2)

I plugged in (x+3) for x and I made y= -1/2(y)
but the graph still doesn't match.

Can you start me like 1/2 way? plz?

4. Originally Posted by greenbee
i don't get it.

I got

y = -(sqrtX+6+(X+3)^2) / (1/2)

I plugged in (x+3) for x and I made y= -1/2(y)
but the graph still doesn't match.

Can you start me like 1/2 way? plz?
First off, please take the time to learn some basic LaTeX. Info can be found at http://www.mathhelpforum.com/math-he...-tutorial.html. It makes reading these equations much less of a chore for such an easy way to make it look nice.

Secondly, you are on the right track! But I see you made a small substitution mistake.

So we decided to replace $x$ with $x+3$, and $y$ with $-2y$. Keep in mind, you need to replace BOTH $x$'s with $(x+3)$. Not just one! Also, watch closely what I do with the 2 on the $y$, remember that to get rid of that 2, we need to either A) Multiply by $\frac{1}{2}$, or divide by 2. Not divide by $\frac{1}{2}$

So before transformation, we have;

$y=\sqrt{2x-x^2}$ After transformations, we get;

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

$y=-\frac{1}{2}\sqrt{2x+6-(x+3)^2}$

If you plot both of these, you will see they match the graphs you provided, check out my following link (Never tried this before with this site, not sure if the link will work, in which case crack out the calculator! EDIT : Copy paste that link and stick it in your URL, it works.)

http://fooplot.com/index.php?&type0=0&type1=0&type2=0&type3=0&type4=0 &y0=sqrt%282x-x^2%29&y1=-%281/2%29sqrt%282x%2B6-%28x%2B3%29^2%29&y2=&y3=&y4=&r0=&r1=&r2=&r3=&r4=&p x0=&px1=&px2=&px3=&px4=&py0=&py1=&py2=&py3=&py4=&s min0=0&smin1=0&smin2=0&smin3=0&smin4=0&smax0=2pi&s max1=2pi&smax2=2pi&smax3=2pi&smax4=2pi&thetamin0=0 &thetamin1=0&thetamin2=0&thetamin3=0&thetamin4=0&t hetamax0=2pi&thetamax1=2pi&thetamax2=2pi&thetamax3 =2pi&thetamax4=2pi&ipw=0&ixmin=-5&ixmax=5&iymin=-3&iymax=3&igx=1&igy=1&igl=1&igs=0&iax=1&ila=1&xmin =-4.36304&xmax=4.52896&ymin=-1.1520000000000001&ymax=1.788

Does this help? Again, read that link I provided in my first post in depth to understand WHY we made these changes.

5. omg thank you so much for the full explanation. I understand! Thank you 1 million zillion times!!!