You can determine this by finding zeros of quadratic function,
[tex](x-1)(x+4)=0\implies x_1=1, x_2=-4[/tex].
Which eliminates first and third option.
Now we have up and down parabolas left.
If we factor [tex](x-1)(x+4)=x^2+4x+\dots[/tex] we find that the coefficient before [tex]x^2[/tex] determines how it is turned, since its positive its an upward turned parabola hence the answer is the last graph.
Hope this helps.
