Find the Horizontal Tangent Line y=2x^3+3x^2-12x+1

Problem

y=2*x3+3*x2−12*x+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)=6*x2+6*x−12

2. Set the derivative to zero because horizontal tangent lines occur where the slope is zero.

6*x2+6*x−12=0

3. Solve for x by first dividing the entire equation by 6 and then factoring the resulting quadratic.

x2+x−2=0

(x+2)*(x−1)=0

x=−2

x=1

4. Find the y-coordinates by substituting the x values back into the original equation y=2*x3+3*x2−12*x+1

y*(−2)=2*(−2)3+3*(−2)2−12*(−2)+1=−16+12+24+1=21

y(1)=2*(1)3+3*(1)2−12*(1)+1=2+3−12+1=−6

5. Write the equations of the horizontal lines using the calculated y values.

y=21

y=−6

Final Answer

y=21,y=−6

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