If I'm given the graph of
y = sqrt(3x-x^2)
how would I stretch that vertically by 3?
this is not so much a transformation problem. what you have is $\displaystyle y = \sqrt{- \bigg( x - \frac 32 \bigg)^2 + \bigg( \frac 32 \bigg)^2}$
or in other words, square both sides and bring the $\displaystyle -(x - 3/2)^2$ over to get
$\displaystyle \bigg( x - \frac 32 \bigg)^2 + y^2 = \bigg( \frac 32 \bigg)^2$
this is therefore, the upper half of a circle centered at $\displaystyle \bigg( \frac 32, 0 \bigg)$ with radius $\displaystyle \frac 32$
so now you would know how to stretch it vertically by 3, right? you would have $\displaystyle y = 3 \sqrt{3x - x^2}$. so the circle looks sort of elliptical. the x-intercepts remain unchanged. but the radius gets distorted vertically
Sorry, but I don't understand.
You say it's not so much a transformation problem, but my book specifically says:
The graph of $\displaystyle y = sqrt(3x - x^2)$ is given. Use transformations to create a function whose graph is as shown.
Let's say I wanted to get it to shift 2 units to the right (or any other transformation). I don't know how to do that with what I have. In the example you showed, I still don't see what to do to transform it.
Thank you for trying to help thus far.
you should have been more specific as to the question. how exactly would i know that is the nature of the question from your first post?
consider a graph $\displaystyle f(x)$ and let $\displaystyle c > 1$ be a constant.
then, $\displaystyle cf(x)$ stretches (vertically) $\displaystyle f(x)$ by a factor $\displaystyle c$ units
$\displaystyle \frac 1c f(x)$ shrinks (vertically) $\displaystyle f(x)$ by a factor of $\displaystyle c$
$\displaystyle f(x + c)$ shifts $\displaystyle f(x)$ to the left $\displaystyle c$ units. ($\displaystyle c > 0$)
$\displaystyle f(x - c)$ shifts $\displaystyle f(x)$ to the right $\displaystyle c$ units. ($\displaystyle c>0$)
$\displaystyle f(cx)$ shrinks $\displaystyle f(x)$ horizontally by a factor of $\displaystyle c$ units. ($\displaystyle c>0$)
$\displaystyle f \left( \frac 1c x \right)$ stretches $\displaystyle f(x)$ horizontally by a factor of $\displaystyle c$ units, c > 0
$\displaystyle f(-x)$ reflects in the y-axis
$\displaystyle -f(x)$ reflects in the x-axis
you should be able to do the transformations you want based on that. so to shift 2 units to the right for example, you want $\displaystyle f(x - 2)$, where $\displaystyle f(x) = \sqrt{3x - x^2}$, so your new graph is that of $\displaystyle f(x - 2) = \sqrt{3(x - 2) - (x - 2)^2}$
I have another question regarding graph transformations.
How does "absolute value" affect the graph of a function?
My book doesn't directly say anything about it, just gives an example. What I understood from the example is that any part of the graph that's below 0 on the y axis is reflected about it, basically changing it from negative to positive. However, when I was doing a problem, the answer in the back of the book didn't seem to be like that.
yes, absolute values make everything positive, which amounts to reflecting (only those parts that are negative) in the x-axis. a more rigorous definition is:
$\displaystyle |x| = \sqrt{x^2} = \left \{ \begin{array}{lr} x & \mbox{ if } x \ge 0 \\ & \\ -x & \mbox{ if } x < 0 \end{array} \right.$