How do you find the vertex of #f(x) = x^2 + 3#?

1 Answer
Apr 25, 2016

#(0, 3)#

Explanation:

The equation of a parabola (with vertical axis) in vertex form can be written:

#f(x) = a(x-h)^2 + k#

where #(h, k)# is the vertex and #a != 0# a constant multiplier.

In our case we find:

#f(x) = 1(x-0)^2+3#

which means that #(h, k) = (0, 3)# and #a=1#

So the vertex is #(0, 3)#

#color(white)()#
Alternatively, note that #x^2 >= 0# with minimum value #0# only when #x=0#. Hence the minimum value of #f(x)# is also when #x=0#. So the #x# coordinate of the vertex must be #0#.

Substituting #x=0# in the formula for #f(x)# gives us #f(0) = 3#.

Hence the vertex is at #(0, 3)#