[Math] $x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$

Question

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?

Answer 1

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.