How do you find the points where the graph of the function #y=(x+3)(x-3)# has horizontal tangents?

1 Answer
Nov 7, 2016

#(0,-9)#

Explanation:

Observe that horizontal tangents occur at a min/max.

You don't need Calculus to answer this question:

# y = (x+3)(x-3) #

The function is a quadratic with roots given by:
# (x+3)(x-3) = 0 => x=+- 3 #
As the coefficient of #x^2# is positive the quadratic will be #uu# shaped and so it will have a minimum, which by symmetry will occur at the midpoint of the roots, i.e when #x=0#

When #x=0=>y=(3)(-3)=-9#

So there is a horizontal tangent at #(0,-9)#

If you should want to use Calculus then:
# y = (x+3)(x-3) => y=x^2-9 #
# :. y' = 2x #
# y' = 0=>2x =0 => x=0 # (as above); so (0,-9) is a critical point

To establish the nature of the critical point we look at the second derivative:
# y'' = 2 > 0 # when #x=0 => # critical point is a minimum, as above.