$f(9-x)=f(9+x)$

as the f(x) satisfies f(9-x)=f(9+x) employs that f(x) is symmetrical about 9 or the 1st derivative which is 2x+a=0 when x=9 therefore value of a=-18 and also graph passes through (0,5) which employs b=5 therefore b-a = 5-(-18)=23

at first put x=0 and f(x)=5,we get b=5.then find a by putting x=9-x and x=9+x and equating we get a=-18.hence b-a=5-(-18)=23

if f(x) = x^2+ax+b satisfies the given function f(9-x)=f(9+x) this we put x= x^2+ax+b in given function and find the value of a. for value of b we pass the given graph y=f(x)= x^2+ax+b from (0,5) thus we get value of b then we subtract b from a and get value.

Note that $b=f(0)$ . Point P gives us the y-intercept of $f(x)$ , which gives $b=5$ .

$f(9-x)=f(9+x)$ tells us that $f(x)$ is symmetrical about $x=9$ . The line of symmetry of a parabola $g(x)=ax^2+bx+c$ is located at $x=\frac{-b}{2a}$ . This gives us $9=\frac{-a}{2}$ , or $a=-18$ .

Finally, we plug in our values for $a$ and $b$ to get $b-a=5-{-18}=\fbox{23}$ .