How do you find the points where the graph of the function #f(x) = -x^2-3x+5# has horizontal tangents?

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Answer 1 · 2018-07-23T14:08:39

#(-3/2, 13/4)#

The given function:

#f(x)=-x^2-3x+5#

Differentiating above function w.r.t. #x# we get the slope of tangent at any point to the given curve as follows

#d/dxf(x)=d/dx(-x^2-3x+5)#

#f'(x)=-2x-3#

The tangent will be horizontal where the slope #f'(x)# of tangent is zero hence

#f'(x)=0#

#-2x-3=0#

#x=-3/2#

setting #x=-3/2# in the function #f(x)#, the y-coordinate of the point is given as follows

#y=f(-3/2)#

#=-(-3/2)^2-3(-3/2)+5#

#=13/4#

The required point is #(-3/2, 13/4)# where tangent is horizontal.