Sketching the graph of S against $\alpha$

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I was trying to solve this problem:
There is a quadratic:
$$f(x)=x^2 - \alpha|x| +2=0 \tag{1}$$

We know that for $|\alpha| < 2\sqrt{2}$, the quadratic would always be smaller than 0.

Let S be the sum of lengths of the intervals in which x satisfies the fact that f(x) <0.

I am able to prove $S<2\alpha$

In this instance, I need to sketch the graph of S against $\alpha$.

Based on the information, we know that the graph would have an asymptote of $$s=2\alpha \tag{2}$$ And the values of S would be $0$ until $\alpha > 2\sqrt{2}$.

When $\alpha>2\sqrt{2}$. I thought it would be a straight line that very close to the asymptote; but, I looked at the solutions given, it is a curve, and it would deviate from the asymptote when $\alpha$ gets large.

May I know why is this so?

Thank you so much you for you guys' replies.