Find the Horizontal Tangent Line y=x^3-3x^2+1

Problem

y=x3−3*x2+1

Solution

1. Find the derivative of the function to determine the slope of the tangent line at any point x

d(y)/d(x)=3*x2−6*x

2. Set the derivative to zero because a horizontal tangent line occurs where the slope is zero.

3*x2−6*x=0

3. Factor the equation to solve for the xvalues where the slope is zero.

3*x*(x−2)=0

4. Solve for x by setting each factor to zero.

x=0

x=2

5. Find the y-coordinates by substituting the xvalues back into the original function y=x3−3*x2+1

y(0)=(0)3−3*(0)2+1=1

y(2)=(2)3−3*(2)2+1=8−12+1=−3

6. Write the equations of the horizontal lines using the calculated yvalues.

y=1

y=−3

Final Answer

y=1,y=−3

Want more problems? Check here!