y = |x²| - 3x - 4. Since the absolute vale of x² is always positive, y will behave the same way as y = x² - 3x - 4.
It's a parabola open upward (a>0) with a axis of symmetry (-b/2a) = 3/2 and y intercept = 4 ( for x=0, y=4).
Moreover x intercepts (roots of the quadratic) → x₁ = 4 and x₂ = -1
All the above correspond to the last graph
Then it's the last graph