Alg 2 question? http://prntscr.com/2ucwq6

Question

tester97 · Answer

@thomaster can you help?

whpalmer4 · Answer

Well, what does the graph of the parent function $f(x) = \frac{1}{x}$ look like?

whpalmer4 · Answer

How does this one differ from that?

tester97 · Answer

Its shifted over to the left a few units and down a few units?

agent0smith · Answer

What makes you think it's shifted left?

tester97 · Answer

sorry im looking at two different sized graphs :/ but it definitely moved down

whpalmer4 · Answer

Yes, it shifted down.  How much?

whpalmer4 · Answer

Look at the behavior of the curve as it goes off the edge at each side.  It's approaching a line there; how much does the line get shifted between the two graphs?

tester97 · Answer

1 unit?

whpalmer4 · Answer

looks like it, yes.  If I have a graph of a function $y = f(x)$, and I want to shift it by 1 unit in the same direction as happened here, what do I do to my function?

whpalmer4 · Answer

4 choices to choose from:  add, subtract, multiply, divide

tester97 · Answer

subtract 1?

whpalmer4 · Answer

how would I write that?

tester97 · Answer

$$y=f(x)-1$$?

whpalmer4 · Answer

very good!

so, if our parent function is $f(x)$, the first step in transforming it was to make it $y = f(x)-1$.

now let's look at that hint...

whpalmer4 · Answer

we have a known point on our graph: $(2,0)$

our function has to be such that $0 = a *f(2) +k$

whpalmer4 · Answer

we have $h=0$ because we didn't shift our graph to the right or to the left. in general, if we have a function $f(x)$, $f(x-h)$ is an exact copy, except shifted $h$ units to the right.  Our graph didn't shift along the x-axis, so that means that we have $h=0$.

Using the hint formula:
$$y = a*f(x-h)+k$$$$y = a*f(x-0)+k$$That's still our parent function as $f(x) = \frac{1}{x}$ so we have
$$0 = a *\frac{1}{x} +k$$

Now insert the numbers we have$$0=a*\frac{1}{2} + k$$
We also know that $k$ is the term that shifts our graph up or down.  We shifted it down by 1 units, so that means $k = -1$.

$$0 = a*\frac{1}{2} -1$$What do you get when you solve that for $a$?

tester97 · Answer

A=2?

whpalmer4 · Answer

yes.  So our equation becomes
$$y = 2*\frac{1}{x}-1$$Let's first check to see if it goes through $(2,0)$:
$$0 = 2*\frac{1}{2}-1$$$$0 = 1-1$$$$0=0\checkmark$$Good, it does!

whpalmer4 · Answer

Now, through the wonders of modern technology, here is a graph of both $$y=f(x) = \frac{1}{2}$$ (in blue)
and $$y = 2f(x-0)-1 = \frac{2}{x-0}-1$$(in purple)

whpalmer4 · Answer

attachment

whpalmer4 · Answer

Notice how it's shifted down by one unit, and reshaped to go through $(2,0)$, but otherwise very similar.

tester97 · Answer

Thank you @whpalmer4 you are very helpful. ^_^ 
$\frak\Huge\color{red}{Thanks~buddy!}$

whpalmer4 · Answer

you're welcome!

do you have another one of these to do?

tester97 · Answer

5 more ;-;

whpalmer4 · Answer

you'll be good at them by the time you're done :-)

what's the next one?  I'll watch over your shoulder...

tester97 · Answer

do you just want me to make a new post? or stay on the same one?

whpalmer4 · Answer

completely up to you.

tester97 · Answer

i'll make a new post xD

tester97 · Answer

getting kinda laggy here ;-;

whpalmer4 · Answer

Sounds good.