How do you find the points where the graph of the function #F(x) = 4x^2 + 4x - 4# has horizontal tangents?

Question

Emilia Aguilar · Accepted Answer

#(-1/2 , -5)#

The function will have horizontal tangents when f'(x) = 0

f(x) = # 4x^2+4x-4 rArr f'(x) = 8x + 4 #

now solve : 8x + 4 = 0 # rArr x = -1/2 #

and #f(-1/2) = 4(-1/2)^2 + 4(-1/2) - 4 = -5 #

hence #(-1/2 , -5)" is the point there is a horizontal tangent " # graph{4x^2+4x-4 [-10, 10, -5, 5]}

Christopher Agresta · Answer

undefined To find the points where the graph of the function F(x) = 4x^2 + 4x - 4 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: F'(x) = 8x + 4.

Next, we set the derivative equal to zero and solve for x: 8x + 4 = 0.

Simplifying the equation, we get: 8x = -4, which gives us x = -1/2.

Therefore, the graph of the function F(x) = 4x^2 + 4x - 4 has horizontal tangents at the point (-1/2, F(-1/2)).

Isabella Ackerman · Answer

undefined To find the points where the graph of the function $ F(x) = 4x^2 + 4x - 4 $ has horizontal tangents, we need to find where the derivative of the function is equal to zero. So, we first find the derivative of $ F(x) $ with respect to $ x $, which is $ F'(x) = 8x + 4 $. Then, we set $ F'(x) = 0 $ and solve for $ x $.

$ 8x + 4 = 0 $

$ 8x = -4 $

$ x = -\frac{4}{8} = -\frac{1}{2} $

So, the point where the graph of $ F(x) $ has a horizontal tangent is $ x = -\frac{1}{2} $.