The quadratic function g(x) has a turning point at (-12,8) where would the quadratic function f(x) = g(4x) have a turning point

For [tex]f(x) = 4x + 1[/tex] and (x) = [tex]g(x)= x^{2} -5,[/tex] find [tex](\frac{g}{f}) (x)[/tex]a. [tex]\frac{x^{2} - 5 }{4x +1 },x[/tex] ≠ [tex]-\frac{1}{4}[/tex]b. x[tex]\frac{4 x +1 }{x^{2} - 5}, x[/tex] ≠ ± [tex]\sqrt[]{5}[/tex]c. [tex]\frac{4x +1}{x^{2} -5}[/tex]d.[tex]\frac{x^{2} -5 }{4x + 1}[/tex]