This is a question coming from an exercices set proposed to students preparing engineering schools competitive exams. Exercice $24$ here : https://www.normalesup.org/~glafon/kaju22/exos_applications.pdf
The question is : graph function $f$ defined by $$ f(x)=\dfrac{x}{2}+\dfrac{1}{3}\sqrt{\left| x^{2}-9\right| }.$$
I wonder which approach to take except plotting a bunch of points and guessing the shape of the curve.
According to Desmos, the graph looks as follows, but how to anticipate this :
I know that applying the absolute value to the quadratic $x^2 -9= ( x-3)(x+3)$ reflects the negative part of the graph by the X axis.
I also see that the square root applied to the quadratic expression flattens its graph , as does the $1/3$ factor.
But my question is : on which ground can one anticipate that the first term ($x/2$) induces a rotation of the graph of the second term ( apparently of $arctan(1/2)$ radians)?
Since $T_1(x)$ is linear and increasing, assume this is the new axis to draw on towards to the positive $x$ direction. While this might look like the illusion of rotation, it is not. Observe that "reflected $x$ section of the graph $(x^2 - 9) < 0$" (as you stated) is "flatter" in $f(x)$ than $T_2(x)$.
This is due to different values of $T_1(x)$ having different effects with that reflected section.
If you want to draw a compound function like $f(x)$ identify all the critical points, and $f(x)$'s direction between the critical points.
We can see that $T'_2(x)$ does not exist at $x = \pm 3$, so its direction would stay the same in $f(x)$ buts image would be translate by $\frac{\pm3}{2}$ units due to the addition of $T_1(x)$.
At $x > 3$, $T_2(x) > 0$ and $T_1(x) > 0$, so $f(x) = T_1(x) + T_2(x) > T_2(x)$
At $f(x) = 0$ , we have to see when $T_1(x) = -T_2(x)$, which is only possible for $x < 0$, since their signs would need to opposite.
$x < x\text{-intercept}$ is left as exercise.