Find the equation of the tangent line for y = x2 + 6 at x = 3 Note: The following examples are part of sections 3.1 - 3.3 homework Find the equation of the tangent line for y = x2 + 6 at x = 3 Solution: The slope of the tangent line at any point is the derivative at that point y' = 2 x At x = 3, we need to find m and y : 3 x = 3 then m =y' (3) = 2(3) = 6 x = 3 then y(3) = (3)2 + 6 = 15 15 Now we want to find the equation of the line that passes the point (x =3, y =15) with m = 6 y = mx + b 15 = (6)(3) + b -3 = b The equation of the tangent line at x = 3 is: y = 6x - 3

The equation of the tangent line at x = 2 is: y = 10x - 15 Find the equation of the tangent line for y = x3 - 2x + 1 at x = 2 Solution: The slope of the tangent line at any point is the derivative at that point y' = 3x2 - 2 At x = 2, we need to find m and y : x = 2 then m =y' (2) = 3(2)2- 2 = 10 5 x = 2 then y(2) = (2)3 -2(2) + 1= 5 Now we want to find the equation of the line that passes the point (x = 2, y = 5) with m = 10 y = mx + b 5 = (10)(2) + b -15 = b The equation of the tangent line at x = 2 is: y = 10x - 15