Question

Help correct please!
Analyze the graph of y = 4x - 4 / x^2 to find the required information.

Answer 1

@helpval22 what required information about the graph do you need to find?

Answer 2

y and x intercepts, vertical and horizontal asymptotes, increasing and decreasing intervals, relative maxima and minima, concave up & down intervals, and the inflection points.
The answer none is also valid. I already did them but I got some wrong, could you help me?

Answer 3

What all have you found so far?

Answer 4

y- intercepts: none, x- intercept: (2,0) ; vertical asympt: x=0 ; horizontal asympt: y=2 ; increasing int: (-infinity, -1) (-1,0) ; decreasing int: (0,4) , ( 4, +infinity)

Answer 5

relative max and min: none? ; concave up int: ( -infinity, -4) , (4, +infinity) ; concave down int: ( -4,4) ; and no inflection points?

Answer 6

How did you determine there is no relative max?

Answer 7

Those I did not found, neither the inflection points. Are the rest correct?

Answer 8

I was just trying to establish a starting point.

Answer 9

I haven't really verified anything yet. I suppose you are forbidden from graphing the equation.

Answer 10

If you graphed it, you would clearly see that there is a relative max.

Answer 11

There's also an inflection point and your x-intercept doesn't seem to be correct either.

Answer 12

Just to verify...

You have posted

\(y = \dfrac{4x - 4}{x^2}\) correct?

Answer 13

Since you have posted linearly, sometimes it is difficult to tell what a student means when they post a fraction in that form.

Answer 14

By literal interpretation what you have posted could be interpreted as

\(y = 4x - \dfrac{4}{x^2}\)

Answer 15

yes, I guess it can. but that is the original equation

Answer 16

I'm sorry, my connection went down for a while...

Answer 17

So of the two graphs I posted, which one is the correct graph?

Answer 18

the first one

Answer 19

Okay. Do you have your notes with you for how to find all these values?

Answer 20

yes, hold on a sec. Ok, I took the first derivative, later the second derivative ( for the inflection points)

Answer 21

the first derivative set to 0 to find the y intercept.
The vertical asympt. where the function is undefined. Horizontal asympt. by getting the limit as x--> infinity

Answer 22

increasing interval, intervals between crtical #s of the first derivative where the derivative is +, for the decreasing interval where it is -negative.
relative maxima points where the derivative goes from + to - . Relative minima where it goes from - to +.

Answer 23

Hmm, as x goes to infinity, y goes to 0

Answer 24

Concave up int: intervals between critical #s of the 2nd derivative where the 2nd derivative is +. Concave down interval where the 2nd derivative is -.

Answer 25

I'm just saying, you have a lot of incorrect values. Tell you what. Why don't you re-post this question and tag some other students like ganeshie8, hartnn, amistre, or myininaya. I'm helping some other students at the moment.