1.

The point at which the tangent to the curve `y=x^(2)-4x` is parallel to x-axis, isA. (0, 4)B. (-2, 4)C. (2,4)D. (2, -4)

Answer» Correct Answer - D
Let `(x_(1),y_(1))` be the required point. Then,
`((dy)/(dx))_((x_(1)","y_(1)))=0rArr 2x_(1)-4=0 rArrx_(1)=2`
Since `(x_(1),y_(1))` lies on `y=x^(2)-4x.`
`therefore y_(1)=x_(1)-4x_(1) rArr y_(1)=4-8= -4`
Hence, the coordinates of the point are (2, -4)


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