Question

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

• A. $(3,-20)$ and $(-1,12)$
• B. $(3,20)$ and $(1,12)$
• C. $(1,-10)$ and $(2,6)$
• D. None of these

Answer 1

Answer: (A)

$(3,-20)$ and $(-1,12)$

Tangent to the curve is parallel to the axis is when slope of the tangent is 0.
$\therefore$ Equation of the curve is
$y=x^3-3x^2-9x+7=0$ ...... $\left(i\right)$
$\therefore \frac{dy}{dx}=3x^2-6x-9$

Now, the tangent is parallel to x-axis, then slope of the tangent is zero or we can say that $\frac{dy}{dx}=0$.
$\Rightarrow 3x^2-6x-9=0$
$\Rightarrow 3(x^2-2x-3)=0$
$\Rightarrow (x-3)(x+1)=0$
$\Rightarrow x=3,-1$

When $x=3$, then from Eq. $\left(i\right),$ we get
$y=(3{)}^3-(3)\cdot (3{)}^2-9\cdot 3+7$
$=27-27-27+7=-20$
When $x=-1,$ then from Eq. $\left(i\right),$ we get
$y=(-1{)}^3-3(-1{)}^2-9(-1)+7$
$=-1-3+9+7=12$
Hence, the points at which the tangent is parallel to x-axis are $(3,-20)$ and $(-1,12)$.