Question

Find the slopes of the tangents of the curve $y=(x+1)(x-3)$ at the points where it cuts the X-axis.

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$

Answer 1

Answer: (A), (B)

The correct options are
A $4$
B $-4$

$f\left(x\right)=(x+1)(x-3)$
Now
$f\left(x\right)=0$
$\Rightarrow x=-1,x=3$
Differentiating $f\left(x\right)$ with respect to x
$\frac{dy}{dx}=x+1+x-3$
$=2x-2$
$=2(x-1)$
Now slope of the tangent at $(h,k)$ will be
${\frac{dy}{dx}}_{h,k}$
Hence slopes of the tangent at $x=-1$ and $x=3$, will be
${\frac{dy}{dx}}_{x=-1}=2(x-1{)}_{x=-1}=-4$
And
${\frac{dy}{dx}}_{x=3}=2(x-1{)}_{x=3}=4$