How do you find the vertex and intercepts for #y=x^2+6#?

1 Answer
Jan 30, 2016

vertex at #(0,6)#
y-intercept at #6#
there is no x-intercept

Explanation:

vertex
The general vertex form is
#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)# with vertex at #(color(red)(a),color(blue)(b))#

Re-writing the given: #y=x^2+6# into explicit vertex form:
#color(white)("XXX")y=color(green)(1)(x-color(red)(0))^2+color(blue)(6)#
with vertex at #((color(red)(0),color(blue)(6))#

y-intercept
The y-intercept is the value of #y# when #x=0#
Setting #x=0# in the given equation:
#color(white)("XXX")y=(0)^2+6=6#
The y-intercept is at #y=6#

x-intercept
The x-intercept is the value of #x# which causes #y=0#
For the given equation #y=x^2+6#
#color(white)("XXX")#if #y=0# then #x^2=-6#
#color(white)("XXX")# but #x >= 0# for all Real values of #x#
Therefore
#color(white)("XXX")#there is no x-intercept.
graph{x^2+6 [-15.54, 12.94, -1.06, 13.18]}