Question

Find points at which the tangent to the curve y = x³ − 3x² − 9x + 7 is parallel to the x axis.

Answer 1

The equation of the given curve is y=x³-3x²-9x+7.

\(\frac{dy}{dx}\)=3x²-6x-9

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

3x²-6x-9=0 ⇒ x²-2x-3=0

=(x-3)(x+1)=0

=x=3 or x=-1

When x = 3, y = (3) ³ − 3 (3) ² − 9 (3) + 7 = 27 − 27 − 27 + 7 = −20. 
When x = −1, y = (−1) ³ − 3 (−1) ² − 9 (−1) + 7 = −1 − 3 + 9 + 7 = 12. 
Hence, the points at which the tangent is parallel to the x-axis are (3, −20) and (−1, 12).