Question

Find points at which the tangent to the curve $y=x^3-3x^2-9x+7$ is parallel to the $x$-axis.

Answer 1

The equation of the given curve is $y=x^3-3x^2-9x+7$ .

$\frac{dy}{dx}=3x^2-6x-9$

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

$\therefore 3x^2-6x-9\Rightarrow x^2-2x-3=0$

$\Rightarrow (x-3)(x+1)=0$

$\Rightarrow x=3orx=-1$

When $x=3,y=(3{)}^3-3(3{)}^2-9\left(3\right)+7=27-27-27+7=-20$.

When $x=1,y=(1{)}^3-3(1{)}^2-9\left(1\right)+7=1-3-9+7=-4$.

Hence, the points at which the tangent is parallel to the x-axis are $(3,-20)$ and $(1,-4)$.