Answer: 3
Step-by-step explanation:
[tex]\displaystyle\ \Large \boldsymbol{} We \ have \ a \ square\ function \\\\ f(x)=ax^2+bx+c \\\\And \ we \ know \\\\ \left[\left \[ {{f(0)=a\cdot 0+b\cdot0+c=0} \atop {f(-4)=a(-4)^2+-4b+c=-24}} \right.=>[/tex] [tex]\displaystyle\ \Large \boldsymbol{} \left[ \ {{c=0} \atop {16a-4b=-24}} \right. =>\boxed{4a-b=-6} \\\\\\ and \ x_0 =-\frac{b}{2a}=1 =>\boxed{ b=-2a} \\\\\\ \left \{ {{4a-b=-6} \atop {b=-2a}} \right. =>4a+2a=-6=> a=-1 \ ; \ b=2 \\\\\\then \ b-a=2-(-1)=\boxed{3}[/tex]
