Question 935851
If {{{f(x)}}} is a linear function, {{{f(x)=mx+b}}} , for some constants {{{m}}} and {{{b}}} .
Then, {{{f(a)=ma+b}}} , and {{{f(0)=m*0+b=0+b=b}}} .
{{{f(x+a)=m(x+a)+b=mx+ma+b=mx+b+ma+b-b=(mx+b)+(ma+b)-b=highlight(f(x)+f(a)-f(0))}}} .