# Graphing Absolute Value Functions

Given: $y = -|2x + 1|-3$

I came up with the graph of...

$1, -7$

$0, -5$

$-1, -3$

$-2, -1$

$-3, 1$

If you were to graph this, it would turn out to be an entirely straight line. This is an absolute value function, so should it not turn at the vertex? What am I missing?

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That function cannot return a positive result. Check your computation for $x=-3$. Your value for $x=-2$ is also wrong. – Brian M. Scott Feb 6 '13 at 2:16

Hint:

$$y=-|2x+1|-3=\begin{cases}-2x-1-3=-2x-4&\text{, if}\;\;\;2x+1\ge 0\Longleftrightarrow x\ge-\frac{1}{2}\\{}\\\;\;\;2x+1-3=\;\;\;2x-2&\text{, if}\;\;\;2x+1<0\Longleftrightarrow x<-\frac{1}{2}\end{cases}$$

so you can see all your values are wrong.

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$$y = -|2x + 1|-3$$

This means: $$y=-|2x+1|-3=\begin{cases}-2x-1-3=-2x-4,&\text{ if}\;\; 2x+1\ge 0\iff x\ge-\frac12\\{}\\ \;\; 2x+1-3=\;\;\;2x-2,&\text{ if}\;\;2x+1<0\iff x<-\frac12\end{cases}$$

Note: $\,y \leq -3\,$ for all values $\,x\,.\,$ When $\,x = -\dfrac12,\; y = -3.\;$

So the "vertex" is at the point $\left(-\frac12, -3\right)$, and as you suggested in your question, the graph of the function $y = -|2x + 1|-3\,$ does, indeed "turn" at this point: when $x \lt -\frac12,\;y$ is increasing (approaching $-3$ from the "left"). When $\,x \gt -\frac12,\;y\,$ is decreasing.

If $f(x)=-|2x+1|-3$ then we can compute points on the line: $$\tag{Points to plot}$$ $$x = 2 \implies f(2) = -|2\cdot 2 +1| - 3 = -|5| - 3 = -8;\tag{2, -8}$$ $$x=1 \implies f(1)=-|2\cdot 1+1|-3=-|3|-3=-6;\tag{1, -6}$$ $$x=0\implies f(0)=-|2\cdot 0+1|-3=-|1|-3 = -4;\quad\tag{0,-4}$$ $$x=-1\implies f(-1)=-|2\cdot(- 1)+1|-3=-|-1|-3=-4\tag{-1, -4}$$ $$x=-2 \implies f(-2)=-|2\cdot(-2)+1|-3=-|-3|-3=-6\tag{-2, -6}$$ $$x = -3\implies f(-3)=-|2\cdot (-3)+1|-3=-|-5|-3=-8\tag{-3, -8}$$

This should give you enough points, starting with the vertex $(-1/2, -3)$ to graph your equation.

Perhaps seeing the graph will help understand what is happening with this function (as you can see, it is NOT a straight line!

$\quad\quad y = -|2x + 1|-3$

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If $f(x)=-|2x+1|-3$ then: $$x=1\to f(1)=-|2\times 1+1|-3=-|3|-3=-6\neq -7\\x=0\to f(0)=-|2\times 0+1|-3=-|1|-3=-4\neq -5\\x=-1\to f(-1)=-|2\times(- 1)+1|-3=-|-1|-3=-4\neq -3\\x=-2\to f(-2)=-|2\times(-2)+1|-3=-|-3|-3=-6\neq -1\\x=1-3\to f(-3)=-|2\times (-3)+1|-3=-|-5|-3=-8\neq -1$$ In fact the function $f$ doesn't meet such these pairs of $(x,y)$ as you noted. This what @Don's answer is trying to show.

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