How do you find the equation of the tangent and normal line to the curve #y=6-x^2# at (2,2)?

1 Answer
Mar 22, 2017

Tangent:

# y = -4x+10 #

Normal:

# y = 1/4x + 3/2 #

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is #-1#

We have:

# y = 6-x^2 #

First let us check that #(2,2)# lies on the curve:

# x=2 => y=6-4=2 #

Then differentiating wrt #x# gives us:

# dy/dx = -2x #

When #x = 2 => dy/dx = -4 #

So the tangent passes through #(2,2)# and has gradient #m_T=-4#, and the normal has gradient #m_N=1/4# so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek are;

Tangent:

# y - 2 = -4(x-2) #
# :. y - 2 = -4x+8 #
# :. y = -4x+10 #

Normal:

# y - 2 = 1/4(x-2) #
# :. y - 2 = 1/4x-1/2 #
# :. y = 1/4x + 3/2 #

We can confirm this solution is correct graphically:
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