Q2 of 149 Page 16

Find the equation of the normal to
y = 2x3 – x2 + 3 at (1, 4).

finding the slope of the tangent by differentiating the curve



m = 4 at (1,4)


normal is perpendicular to tangent so, m1m2 = – 1



equation of normal is given by y – y1 = m(normal)(x – x1)



x + 4y = 17


More from this chapter

All 149 →
21

Find the point on the curve y = x3 where the Slope of the tangent is equal to x – coordinate of the point.

 1

Find the equation of the tangent to the curve at the point (a2/4, a2/4)

3

Find the equation of the tangent and the normal to the following curves at the indicated points:

y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)

 3

Find the equation of the tangent and the normal to the following curves at the indicated points:

y = x4 – 6x3 + 13x2 – 10x + 5 at x = 1 y = 3