Find the point on the curve `y=2x^2-6x-4` at which the tangent is parallel to the x-axis.

Answer

`(3 / 2,-17 / 2)`

Explanation

\begin{aligned}
&y=2 x^{2}-6 x-4 \\
&y^{\prime}=4 x-6=0 \\
&x=3 / 2 \\
&y=2(3 / 2)^{2}-6(3 / 2)-4 \\
&=2(9 / 4)-9-4 \\
&=(9 / 2)-13 \\
&=(9-26) / 2 \\
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