Question

The points 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

The correct option is E

$\left(\frac{3}{2},-\frac{17}{2}\right)$

The explanation for the correct answer.

Solve for the points on the curve $y=2x^2-6x-4$

$y=2x^2-6x-4$

Find the first derivative and substitute it equal to zero as tangent is parallel to $x-axis$

$y\text{'}=4x–6=0$

Find the value of $x$

$$\begin{gathered} \Rightarrow x=\frac{3}{2} \\ andpointy \\ y=2{\left(\frac{3}{2}\right)}^2-6\left(\frac{3}{2}\right)-4 \\ =2\left(\frac{9}{4}\right)-9-4 \\ =\frac{9}{2}-13=\left(-\frac{17}{2}\right) \end{gathered}$$

So, the points are $\left(\frac{3}{2},-\frac{17}{2}\right)$

Hence, option(E) is the correct answer.