InterviewSolution
Saved Bookmarks
| 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) |
|