Browse Questions

# The point on the curve $y=3x^{2}-6x-4$ at which the tangent is parallel to the $x$ -axis is

$\begin {array}{1 1}(1)\bigg(\frac{5}{2},\frac{-17}{2}\bigg)&(2)\bigg(\frac{-5}{2},\frac{-17}{2}\bigg)\\(3)\bigg(\frac{-5}{2},\frac{17}{2}\bigg)&(4)\bigg(\frac{3}{2},\frac{-17}{2}\bigg)\end{array}$

$y=2x^2-6x-4$
$\large\frac{dy}{dx}$$=4x-6 Let (x_1,y_1) be the point on the curve at which the tangent is parallel to x-axis \large\frac{dy}{dx}$$/(x_1,y_1)=4x_1-6$
Since the tangent is parallel to x-axis
$\large\frac{dy}{dx} $$/(x_1,y_1)=0 4x_1-6=0=> x_1=\large\frac{3}{2} (x_1,y_1) is a point on the curve \therefore y_1=2x_1^2-6x_1-4 \qquad= 2 \times \large\frac{9}{4}$$ -6 \times \large\frac{3}{2}$$-4 \qquad= \large\frac{9}{2}-\frac{18}{2}$$-4$
$\qquad= \large\frac{9-18-8}{2}$
$\qquad= \large\frac{-17}{2}$
The point is $\bigg(\frac{3}{2},\frac{-17}{2}\bigg)$
Hence 4 is the correct answer