Given that f(x) = x2 − 4x − 3 and g(x) = the quantity of x plus three, over four , solve for f(g(x)) when x = 9? −6 3 6 9

#"to evaluate "f(g(x))" substitute "x=g(x)" into "f(x)#

#g(x)=(x+3)/4#

#f(g(x))=f((x+3)/4)#

#=((x+3)/4)^2-cancel(4)((x+3)/cancel(4))-3#

#=((x+3)^2)/16-(x+3)-3#

#f(g(color(red)(9)))=((color(red)(9)+3)^2)/16-(color(red)(9)+3)-3#

#color(white)(xxxxx)=144/16-12-3#

#color(white)(xxxxx)=9-15=-6#

Given that #g(color(magenta)(x))=(color(magenta)(x)+3)/4 color(white)("ddd")->color(white)("ddd") g(color(magenta)(9)) = (color(magenta)(9)+3)/4#

So wherever there is an #x# put a #color(magenta)(9)# instead.

#g(9)=12/4=color(green)(3)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that #f(x)= x^2-4x-3#

So wherever there is a #g(x)# put a #color(green)(3)#

#f(g(x)) =f(color(white)("d")ubrace(g(color(magenta)(9)))color(white)("d"))#
#color(white)("dddddddddddd")darr#
#f(g(x)) = f(color(white)("d.d")color(green)(3)color(white)("dd."))#

#color(brown)(f(color(green)(3))=color(green)(3)^2color(white)("dd")ubrace(-4(color(green)(3))-3))#
#color(brown)(color(white)("ddddddddddd.d")darr#
#f(3)=+9color(white)("d")-15color(white)("ddd")=-6#