Question 410515
(g-f)(x)
= g(x) - f(x) 
= (x^2 + 3) - (5x - 4) 
= x^2 + 3 - 5x + 4
= x^2 - 5x + 7


(fg)(x)
= f(x) * g(x)
= (5x-4) * (x^2 + 3)
= 5x^3 + 15x - 4x^2 - 12
= 5x^3 - 4x^2 + 15x - 12


(g/f)(x)
= g(x) / f(x)
= (x^2 + 3) / (5x-4)
the denominator = 0 when 5x-4 = 0
this occurs when x = 4/5
the function is undefined when the denominator is equal to 0
the domain is therefore all real values of x except when x = 4/5


the graph of this equation looks like this:


{{{graph(600,600,-5,5,-5,5,(x^2+3) / (5x-4))}}}


if you draw a vertical line at x = 4/5, you will see that the graph of the function approaches that line but never touches it.  that means that the line x = 4/5 is an asymptote of the function which also means that that the function is undefined at x = 4/5.