Answer to Question #304845 in Calculus for jin

Question

Find the equations of the tangent and the normal of y = 3x²-2x+1 at point (1,2).

Answer 1

$$y'=(3x^2-2x+1)'=6x-2$$

Point (1,2)

$$slope_1=m_1=y'(1)=6(1)-2=4$$

$$slope_2=m_2=-\dfrac{1}{y'(1)}=-\dfrac{1}{4}$$

The equation of the tangent line is

$$y=4(x-1)+2$$

$$y=4x-2$$

The equation of the normal line is

$$y=-\dfrac{1}{4}(x-1)+2$$

$$y=-\dfrac{1}{4}x+\dfrac{9}{4}$$