00:01
So in this problem, first we find slope of tangent that is d .y upon dx.
00:06
So dy upon dx would be 3x square minus 6x plus 2.
00:13
Now we find points where curve intersect x -axis.
00:19
So for that on x -axis, y equal to always 0.
00:23
So i will be putting y equal to 0 here.
00:25
So to find point xq minus 3x square plus 2x equal to 0.
00:35
Now we simplify this by taking x common.
00:40
So this would be x square minus 3x plus 2 equal to 0.
00:46
Now we rectize the bracket that will give us x minus 1 x minus 2 equal to 0.
00:55
So points on the x -axis would be from here x equal to 0, 1 and 2.
01:06
So points would be 0, 0, 1 comma 0 and 2.
01:18
Now we will be finding slope on these points.
01:22
So slope m1 would be d -y upon d x at 0 .0...