$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as

$(0, -7)$; no $x$ intercepts.

I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3}$ and $2+\frac{1}{3}$
My working:
$-3|x-2|-1=0$
$-3|x-2|=1$
$|x-2|=-\frac{1}{3}$
Then solve for both the negative and positive value of $\frac{1}{3}$:
Positive version:
$x-2=\frac{1}{3}$
$x=2+\frac{1}{3}$
Negative version:
$x-2=-\frac{1}{3}$
$x = 2-\frac{1}{3}$
How can I arrive at "$(0, -7)$; no x intercepts"? Where did I go wrong in my understanding?
 A: The range of $y=|x|$ is $\{y\,|\,y\in\Bbb R, y\geq 0\}$.  Therefore, 
$$|x-2|=-\frac13$$
has no real solution.

The curves do not intersect.
A: For $$x\geq 2$$ we get $$3(x-2)-1=3x-6-1=3x-7$$ .For $$x<2$$ we have $$3(2-x)-1=-3x+5$$
A: The graph of the function $f(x) = -3|x - 2| - 1$ has vertex $(2, -1)$ and opens downward since the coefficient of $|x - 2|$ is negative.  That tells you that the graph of the function cannot intersect the $x$-axis.
Remember that $|x|$ means the distance of the number $x$ from $0$.  Therefore, $|x - 2|$ means the distance of the number $x - 2$ from $0$.  However, a distance cannot be negative.  
To solve for the $x$-intercepts, you set $f(x) = 0$, which yields
\begin{align*}
f(x) & = 0\\
-3|x - 2| - 1 & = 0\\
-3|x - 2| & = 1\\
|x - 2| & = -\frac{1}{3}
\end{align*}
which is impossible since $|x - 2| \geq 0$ for every real number $x$.  Hence, there are no $x$-intercepts.  
We know that $f(0) = -3|0 - 2| - 1 = -3|-2| - 1 = -3(2) - 1 = -6 - 1 = -7$.  By symmetry, $f(4) = -7$, as you can check.

A: $$|x-2|=-\frac{1}{3}$$
Two cases:
1) If $x\ge2$, the equation becomes $x-2 = -\frac{1}{3}$, which has no solution. 
2) If $x<2$, the equation becomes $-(x-2 )=-\frac{1}{3}$, which has no solution. 
Thus, there is no x intercepts. (0,-7) is the y intercept as seen from  $f(0)=-7$.
