How do you graph #y=1/2(x-3)^2+5#?

1 Answer
Sep 20, 2015

Determine the axis of symmetry and the vertex. Determine points on both sides of the axis of symmetry. Sketch a parabola through the points. Do not connect the dots.

Explanation:

#y=1/2(x-3)^2+5# is in vertex form, #y=a(x-h)^2+k#, where #a=1/2, h=3, and k=5#.

The vertex is #(h,k)#, which is #(3,5)#. The axis of symmetry is #x=h=3#

Substitute several values for #x# on both sides of the axis of symmetry to find points on the parabola.

#x=6,# #y=19/2#

#x=5,# #y=7#

#x=4,# # y=11/2#

#x=3,# #y=5# (vertex)

#x=2,# #y=11/2#

#x=1,# #y=7#

#x=0,# #y=19/2#

Plot the points and sketch a curved parabola. Do not connect the dots.

graph{y=1/2(x-3)^2+5 [-16.49, 15.53, 0, 16.01]}