How do you find the equation of a line tangent to the function #y=x^2-2# at x=0?

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Answer 1 · 2016-11-02T19:58:24

# y=-2 #

If #y=x^2-2# then #dy/dx=2x#

When #x=0 #
# => y=-2 #
# => dy/dx=0 # (ie we are at a critical point!)

so the tangent passes through #(0,-2)# and has gradient #m_T=0#

Using #y-y_1=m(x-x_1)# the equation of the tangent is:

# y-(-2)=0 #
# :.y=-2 #

It should also be obvious that #dy/dx=0# means the point corresponds to a critical point and consequently a horizontal tangent.

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