<s>[INST] $x$ 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?
 [/INST] $$|x-2|=-\frac{1}{3}$$
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  $f(0)=-7$.
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