Question

State a function that will result in the same graph:
a) abs(2x-3)
b)abs(-(x+3)^2-2)
c)abs(x^2+1)
d)abs((x+2)^2-4)

Answer 1

Where is the graph?

Answer 2

shoot just a sec

Answer 3

This is graph a and graph b

Answer 4

and this is graph c and d - if you need the points for any i can give those too

Answer 5

Just to make sure I'm understanding the question correctly... you were given these four functions, and you plotted them. Now you have to give another expression for the same functions?

Answer 6

Yep!

Answer 7

Okay, (b) and (c) are the simplest ones.
\[\begin{align*}(b)~~\left|-(x+3)^2-2\right|&=\left|(-1)\big((x+3)^2+2\big)\right|\\
&=|-1|\left|(x+3)^2+2\right|\\
&=\left|(x+3)^2+2\right|
\end{align*}\]
Since \((x+3)^2\) will always be positive, adding 2 to it will not change this fact, so you can drop the absolute value bars altogether, so that
\[\left|-(x+3)^2-2\right|=(x+3)^2+2\]
You can tell this is the case by looking at the graph you drew.

(c) can be explained identically. Since \(x^2\) is positive, so must \(x^2+1\) also be positive, so \(|x^2+1|=x^2+1\), also agreeing with your graph.

Answer 8

(a) and (d) will require a piecewise definition. Recall that
\[|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\]
So, examining (a),
\[|2x-3|=\begin{cases}2x-3&\text{for }2x-3\ge0\\-(2x-3)&\text{for }2x-3<0\end{cases}\]
Rewriting a bit, you have
\[|2x-3|=\begin{cases}2x-3&\text{for }x\ge\dfrac{3}{2}\\\\-2x+3&\text{for }x<\dfrac{3}{2}\end{cases}\]

Answer 9

(d) is somewhat trickier:
\[\left|(x+2)^2-4\right|=\begin{cases}(x+2)^2-4&\text{for }(x+2)^2-4\ge0\\-\left((x+2)^2-4\right)&\text{for }(x+2)^2-4<0\end{cases}\]
To simplify the conditions so that they're in terms of \(x\) only, let's consider for which values \((x+2)^2-4\) is positive and for which it's negative. To do so, we'll consider the roots of the expression, then see what happens between and around the roots:
\[(x+2)^2-4=0~~\iff~~(x+2)^2=4~~\iff~~x+2=\pm2~~\iff~~x=0,~x=-4\]
For when \(x<-4\), let's pick \(x=-5\). This gives
\[(-5+2)^2-4=9-4=5>0\]
For when \(-4<x<0\), let's use \(x=-1\), then
\[(-1+2)^2-4=1-4=-3<0\]
For when \(x>0\), let's check with \(x=1\), so that
\[(1+2)^2-4=9-4=5>0\]
Basically, what we've determined is that \((x+2)^2-4\) is positive when \(x<-4\) and when \(x>0\), and negative when \(-4<x<0\). In terms of the piecewise function, it'd be
\[\left|(x+2)^2-4\right|=\begin{cases}(x+2)^2-4&\text{for }x\le-4\text{ and }x\ge0\\-\left((x+2)^2-4\right)&\text{for }-4<x<0\end{cases}\]
Where you put the \(<\) or \(\le\) doesn't matter in this case, it's up to you.

Answer 10

Holy, thank you SO much. You basically just taught me a whole section of my textbook that I didn't understand

Answer 11

You're welcome!

Answer 12

So b and c will work pretty much the same as that??

Answer 13

Yes, you could use the absolute value definition, or you could do what I did and notice that since that whatever's inside the absolute value bars is always positive, the absolute value of some positive number is also positive.

Answer 14

Sweet! Thank you kindly :)