Question

The slope of the tangents to the curve $y=(x+1)(x-3)$ at the points where it crosses x - axis are

• A. $\pm 2$
• B. $\pm 3$
• C. $\pm 4$
• D. None of these

Answer 1

Answer: (C)

$\pm 4$

$y=(x+1)(x-3)..\left(1\right)$

Substitute $y=0$ to get the point of intersection of this curve with $x-axis$

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

$\Rightarrow x=-1,3$

So the points are $(-1,0)$ and $(3,0)$

Now differentiating eq. (1), we get

$\frac{dy}{dx}=(x-3)+(x+1)=2x-2$

$\Rightarrow \left(\frac{dy}{dx}\right)=2(x-1)$

Thus slope at $(-1,0)$ is $={\left(\frac{dy}{dx}\right)}_{(-1,0)}=-4$

and at $(3,0)$ is $={\left(\frac{dy}{dx}\right)}_{(3,0)}=4$

Hence, option 'C' is correct.