How do you find the important points to graph #f(x)= -x^2-4x#?

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Answer 1 · 2018-01-18T17:13:48

Check below for detail examination of the function.

#f(x)=-x^2-4x# ,
#D_f=RR#

#f(x)=0 <=> -x^2-4x=0 <=> x^2+4x=0 <=> x(x+4)=0 <=> (x=0, x=-4)#

#f'(x)=-2x-4=-2(x+2)#

#f'(x)=0# #<=> x=-2#

#f''(x)=-2<0#
#x##in##RR#

#f# has global maximum at #x_0=-2# , #f(-2)=4#

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#lim_(xrarr-oo)f(x)=lim_(xrarr-oo)(-x^2-4x)=lim_(xrarr-oo)(-x^2)=-oo#

#f(-2)=4#

#lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(-x^2-4x)=lim_(xrarr+oo)(-x^2)=-oo#

#f# is defined in #RR# so it doesn't have any vertical asymptotes.

#lim_(xrarr-oo)f(x)/x=lim_(xrarr-oo)(-x^2-4x)/x=lim_(xrarr-oo)(-x^2/x)=+oo#

enter image source here