Question

The point on the curve $y=2x^2-6x-4$ at which the tangent is parallel to the $x-$ axis, is:

• A. $\left(\frac{3}{2},\frac{13}{2}\right)$
• B. $\left(-\frac{5}{2},-\frac{17}{2}\right)$
• C. $\left(\frac{3}{2},\frac{17}{2}\right)$
• D. $(0,-4)$
• E. $\left(\frac{3}{2},-\frac{17}{2}\right)$

Answer 1

Answer: (E)

Tangent to any curve which is parallel to $x-$ axis, if $\left(\frac{dy}{dx}\right)=0$ Given $y=2x^2-6x-4$ $\Rightarrow$ $\frac{dy}{dx}=4x-6=0\Rightarrow x=\frac{6}{4}=\frac{3}{2}$ $\Rightarrow$ $y=2.\frac{9}{4}-6.\frac{3}{2}-4$ $=\frac{9}{2}-9-4=\frac{9-18-8}{2}=-\frac{17}{2}$ $\therefore$ Required point is $\left(\frac{3}{2},-\frac{17}{2}\right)$ .