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?

May 1, 2015

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

Therefore, we just have to find $a$ if
$0 = \frac{1}{2} {\left(a - 8\right)}^{2} + 2$

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

(B) Axis of symmetry of a canonical parabola $y = {x}^{2}$ is the Y-axis.
The parabola $y = \frac{1}{2} {x}^{2}$ differs from a canonical one only by a multiplier $\frac{1}{2}$, 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 $y = F \left(x\right)$ shifts the graph to the right by the value of this constant.
Therefore, the axis of symmetry of $y = \frac{1}{2} {\left(x - 8\right)}^{2}$ 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 $\frac{1}{2} {\left(x - 8\right)}^{2} + 2$ has a line $x = 8$ as the axis of symmetry.

(C) The minimum point of a canonical parabola $y = {x}^{2}$ is a point $\left(0 , 0\right)$.
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 $\left(8 , 0\right)$.
After that we add $2$ to a function, which shifts the graph upwards by $2$, so the minimum point shifts to $\left(8 , 2\right)$.

(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:
$y = \frac{1}{2} {\left(0 - 8\right)}^{2} + 2$
from which we derive $y = 34$.
Therefore, a point $\left(0 , 34\right)$ 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 $\left(8 , 2\right)$ and the Y-intercepts is $34$.