What is the vertex if g(x) = 8x2 - 64x?

Find and classify the global extrema of the following function:

f(x) = 8 x^2 - 64 x

Find the critical points of f(x):

Compute the critical points of 8 x^2 - 64 x

To find all critical points, first compute f'(x):

d/( dx)(8 x^2 - 64 x) = 16 x - 64

= 16 (x - 4):

f'(x) = 16 (x - 4)

Solving 16 (x - 4) = 0 yields x = 4:

x = 4

f'(x) exists everywhere:

16 (x - 4) exists everywhere

The only critical point of 8 x^2 - 64 x is at x = 4:

x = 4

The domain of 8 x^2 - 64 x is R:

The endpoints of R are x = -∞ and ∞

Evaluate 8 x^2 - 64 x at x = -∞, 4 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

4 | -128

∞ | ∞

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

4 | -128 | global min

∞ | ∞ | global max

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

4 | -128 | global min

f(x) = 8 x^2 - 64 x has one global minimum:

 f(x) has a global minimum at x = 4

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

glide reflection.

its mainly because the pattern will be infinite and will never end. it will look like a gilder or like a filpbook when set to motion.