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

Question

Christopher Allers · Accepted Answer

#(-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.

Cameron Alexander · Answer

undefined To find the points where the graph of the function f(x) = -x^2-3x+5 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of f(x) by applying the power rule and the sum rule of differentiation. The derivative of -x^2 is -2x, the derivative of -3x is -3, and the derivative of 5 is 0. Therefore, the derivative of f(x) is -2x - 3.

Next, we set the derivative equal to zero and solve for x: -2x - 3 = 0. Adding 3 to both sides gives -2x = 3, and dividing by -2 gives x = -3/2.

So, the graph of the function f(x) = -x^2-3x+5 has horizontal tangents at the point (-3/2, f(-3/2)).