Tangent line using basic differentation

Question

If , find .   Use this to find the equation of the tangent line to the parabola  at the point . The equation of this tangent line can be written in the form  where  is:  and where  is:   

Answer 1

F(x) = 2X² - 2X + 6 → F^'(X) = 4X - 2 → F^'(- 5) = 4(- 5) - 2 = - 22

Equation of tangent line at ( X1,Y1) is: Y - Y1 = m ( X - X1)

In this problem X1 = - 5 , Y1 = 66 and slope = m = F^'(- 5) = - 22 → Y - 66 = - 22 [X - (-5)]

→ Y = -22X + (-22) (5) + 66 → Y = -22X - 44 (Y = mX + b)

At X = 0 → Y = - 22(0) - 44 or b = - 44 (Y intercept)