The slope of tangent on the curve y=6 x^2-x-12, at the point where cu…

The correct answer is Option (1) → 17

at positive x axis $y=0,x>0$

so $6x^2-x-12=0$

$6x^2-9x+8x-12=0$

$3x(2x-3)+4(2x-3)=0$

$x=\frac{-4}{3},x=\frac{3}{2}$

$x=\frac{3}{2}$ as $x>0$

so $\frac{dy}{dx}=12x-1$

$\left.\frac{dy}{dx}\right]_{x=\frac{3}{2}}=18-1=17$