Question

Let $y=x^3-6x^2+9x+1$ be the equation of a curve, then the $x$ -intercept of the tangent to this curve whose slope is least, is

Answer 1

Answer: 3

$\frac{dy}{dx}=3\left(x^2-4x+3\right)=3\left[{\left(x-2\right)}^2-1\right]\ge -3$
So, minimum value is $-3$ when $x=2$
at $x=2,y=3$
Hence, equation of tangent is $3x+y=9$
So, $x$ -intercept $=3$