How do you find x-intercepts, axis of symmetry, maximum or minimum point, and y-intercept for the equation #y=1/2(x-8)^2+2#?

1 Answer
May 1, 2015

(A) X-intercept is a point where a graph intersects the X-axis, that is a point with coordinates #(a,0)# that satisfies our equation when we substitute #x=a# and #y=0#.

Therefore, we just have to find #a# if
#0=1/2(a-8)^2+2#

This equation, obviously has no real solutions because #1/2(a-8)^2 >= 0# because it's a square of some number multiplied by a positive constant, and, therefore, #1/2(a-8)^2+2 > 0# and there are no such values of #a# when the expression #1/2(x-8)^2+2# equals to #0#.

(B) Axis of symmetry of a canonical parabola #y=x^2# is the Y-axis.
The parabola #y=1/2x^2# differs from a canonical one only by a multiplier #1/2#, which just squeezes the parabola towards the X-axis by a factor of #2# without changing its axis of symmetry.

Subtracting a positive constant from an argument #x# in the equation of any function #y=F(x)# shifts the graph to the right by the value of this constant.
Therefore, the axis of symmetry of #y=1/2(x-8)^2# is a straight line parallel to the Y-axis and intersecting the X-axis at point #x=8#. The equation of this line is #x=8# (independent of #y#).

Adding #2# to a function shifts the graph upwards without changing it's axis of symmetry. So, parabola #1/2(x-8)^2+2# has a line #x=8# as the axis of symmetry.

(C) The minimum point of a canonical parabola #y=x^2# is a point #(0,0)#.
After squeezing it by a factor of #2# the minimum point does not change its position.
After subtracting #8# from the argument the whole graph shifts to the right by #8#, and the minimum point shifts by #8# as well, taking a position #(8,0)#.
After that we add #2# to a function, which shifts the graph upwards by #2#, so the minimum point shifts to #(8,2)#.

(D) Y-intercept is a point on the Y-axis where the graph intersects this axis. This is a value of a function when an argument equals to zero.
Substitute #x=0# in the function:
#y=1/2(0-8)^2+2#
from which we derive #y=34#.
Therefore, a point #(0,34)# is a Y-intercepts.


Let's illustrate it graphically:
graph{1/2(x-8)^2+2 [-1, 20, -10, 40]}
On the above graph you see that there is no X-intercepts, the axis of symmetry is a line #x=8#, the minimum point is #(8,2)# and the Y-intercepts is #34#.