How do you find the equation of the line tangent to #y=x^4# at (-2,16)?

Question

6152 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T16:10:39

# y = -32x-48 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

so If # y= x^4 # then differentiating wrt #x# gives us:

# \ \ \ \ \ dy/dx = 4x^3 #

When #x = -2 => y = 16 # (so #(-2,16)# lies on the curve)
and # dy/dx = 4(-8)=-32#

So the tangent passes through #(-2,16)# and has gradient #-32#, so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# y-(16) = -32(x-(-2)) #
# :. y-16 = -32x-64 #
# :. y = -32x-48 #

We can confirm this solution is correct graphically:
enter image source here