0:00
therefore c is equals 2 34 these implies
0:03
that the equation of the tangent
0:05
straight line that has the point of
0:06
tangency at x= 22 we also have the
0:09
coordinates of the point of tangency on
0:11
the curve whereby we substituted the x
0:13
coordinate into the original function gx
0:16
and that will be giving us the
0:17
y-coordinate of the point where the
0:18
tangent touches the curve that means we
0:20
can find the equation of the tangent
0:23
line now the equation of a line is y = 2
0:26
mx + c where m is the gradient and c are
0:28
the y intercept or the constant we know
0:30
the gradient is minus 8 and the
0:32
coordinates are 2 as the x coordinates
0:34
and 18 as the y-coordinate we need to
0:36
find the value of the constant c the y
0:39
intercept of the tangent line so let's
0:41
go ahead and substitute the values into
0:43
this equation y = mx + c so we have 18 =
0:47
to8 * 2 + c8 * 2 is equals to 16 and
0:51
then transpose -16 to the left of the
0:54
equal sign this implies 18 + 16 which is
0:57
equals to 34 therefore C is equals to 34
1:01
these implies that the equation of the
1:03
tangent straight line that has the point
1:05
of tangency at x= 22 is then given by
1:08
the following equation y =8x + 34
