The slope of tangent on the curve $y=6 x^2-x-12$, at the point where curve meets positive x-axis is _________. |
17 -17 $\frac{3}{2}$ 19 |
17 |
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 $\frac{dy}{dx}=12x-1$ $\left.\frac{dy}{dx}\right]_{x=\frac{3}{2}}=18-1=17$ |