# 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##?

(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##.