For #f(x)=-2x^2-5x#
To find the max/min value we need to obtain the form
#a(x-h)^2+k#.
Where #a# is the coefficient of #x^2#, #h# is the axis of symmetry and #k# is the max/min value.
Factor out the coefficient of #x^2#.
#-2(x^2+5/2x)#
Add the square of half the coefficient of #x# inside the brackets and subtract the square of half the coefficient of #x# outside of the brackets. Remember to multiply the square of half the coefficient by #-2# when subtracting it outside the brackets, since we factored this out at the beginning.
#-2(x^2+5/2x+(5/4)^2)-(-2)(5/4)^2#
Make into the square of a binomial and simplify:
#-2(x+5/4)^2+25/8#
Since the coefficient of #x^2# is negative the parabola will be of this form #nnn#, so #k# is a maximum value. #k=25/8#
#f(x)# is a polynomial so has domain #RR#
#{ x in RR}#
For end behaviour we only need to look at the leading term and degree.
#:.#
#-2x^2#
as #x->+-oo# ,#color(white)(888)-2x^2->-oo#
So range is:
#{y in RR : -oo < y <= 25/8 }#
For #g(x)# This is a line, so it has domain and range #RR#
#{x in RR }#
#{y in RR }#
#f(g(x))=-2(g(x))^2-5(g(x))=-2(3x+2)^2-15x-10=-18x^2+9x-18#
#-18(x-1/4)^2-143/8# ,as before this is a maximum value.
Domain:
#{ x in RR }#
Range:
#{y in RR : -oo< y <= -143/8}#
