Quiz 3 Quiz 3                              Friday, March 30, 2001



NAME:

Student Number:------            No.----           Marks ----






(20 Marks) Find the equation of the tangent line to y = [(3x-5)/(x2+7)] at the point (0, -5/7).

Solution.


y
=
(3x-5)(x2+7)-(3x-5)(x2+7)
(x2+7)2
=
(3(x2+7)-(3x-5)(2x)
(x2+7)2
=
-3x2+10x +21
(x2+7)2

Hence we have y(0) = [21/47] = [3/7].

The equation of the tangent line is


y = (-5/7)+y(0)(x-0) = -5/7+(3/7)x

or


7y-3x+5 = 0.




File translated from TEX by TTH, version 2.79.
On 2 Apr 2001, 14:18.