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##.

Leave a Comment

Your email address will not be published. Required fields are marked *

Is this question part of your Assignment?

Get expert help

Girl in a jacket


We are a team of academic consultants with extensive experience in writing academic papers for college students in the US, Canada, UK, AU, and other parts of the world.

We help students with both technical and non-technical assignments across all majors & academic disciplines.

Unlike what our name suggests, we research and draft everything word for word. We do not use AI or any rewriting tool! We provide Turnitin reports for AI & Turnitin alongside every paper.

Need help? Send us your assignment now!

description here description here description here