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help me with a graph!

Blogs > Raithed
Post a Reply
Raithed
Profile Blog Joined May 2007
China7078 Posts
Last Edited: 2008-06-04 17:21:23
June 04 2008 17:18 GMT
#1
y = x^2 - 1 / x^3

1. x-int:
2. y-int:
3. vertical asymp:
4. horiz. asymp:
5. relative extrema:
6. points of inflection:

not asking to finish my hw, just to start it up!

y = x^2 - 1 / x^3 looks like:
[image loading]



1. for 1 and 2, i have no idea what it is asking. what should one do first?
2. --
3. for vertical asym; isnt there a rule that states if the power of the numerator is greater than the power of the denominator, there is a vertical asymp. or something?
4. by limit standards, it goes to 0?
5. can someone give me a hint on how to start it?
6. i know an inflection point is basically a change of curves, does this mean i should take the derivative? (quotient rule)

gwho
Profile Blog Joined January 2008
United States632 Posts
June 04 2008 17:28 GMT
#2
my advice is to read your book. lol. asic stuff. but what the hey, i'll give you a few pointers... maybe even do your hw for u


the solution (graph) never touches the x or y intercept. in this case you say there is no x or y intercept.

but notice how as x goes to 0 or infinity, they "approach" a certain value. these are asymptotes and you describe them with a line. as x goes to zero, the y value approaches infinity. the asymptote coincides with the y axis. so your vertical asmptote is x=0. and i'm sure u can now handle do the horizontal asymptote.

relative extreme as i remember would be humps or other extrema that aren't the maximum. in this case, your maximums and minimums are all the greatest ones. so you have absolute extremes of infinity and negative infinity at x --> 0.

inflection are when the 2nd derivative changes signs. so normally u do take the derivative... twice. but of course in this one, it's not an inflection but a discontinuity at x=0 so there are no inflection points. if anything it would be x=0, but maybe they call it something else i can't remember.


Raithed
Profile Blog Joined May 2007
China7078 Posts
June 04 2008 17:45 GMT
#3
okay so i just basically do y'' for the inflection point?
gwho
Profile Blog Joined January 2008
United States632 Posts
June 04 2008 19:31 GMT
#4
what? a y? i don't thin kthere are any inflection points. i would just write x=0 as a discontinuity. but it's definitely not an inflection point
Raithed
Profile Blog Joined May 2007
China7078 Posts
June 04 2008 19:37 GMT
#5
when i mean y" i mean f''(x) sorry been changing y and f(x) a lot lately.
Day[9]
Profile Blog Joined April 2003
United States7366 Posts
June 04 2008 19:38 GMT
#6
first of all, SOLVE FOR EVERYTHING ALGEBRAICALLY. do NOTHING visually. only use the graph as a reality check. (ie, if you find one x-intercept, but there are clearly three, you know you did something wrong). Do solve for each one algebraically, do the following.

1. x-int:
This is asking for the x-intercept AKA the points where the graph touches the x-axis. At points where the graph touches the x-axis, we know that x=0 (by definition). Consequently, to solve for the x-intercepts, you simply plug x=0 into your equation and solve for y. Your x-intercepts will be coordinates of the form (0,y1) (0,y2) (0,y3)... where y1 is the first x-intercept, y2 is the second x-intercept y3... etc etc

2. y-int:
Just as before, this is asking for the y-intercept AKA the points where the graph touches the y-axis. Again, to find this, set y=0 and solve for x in your graph equation. Your answers will be coordinate points again.

3. vertical asymp:
A vertical asymptote is a point at which the graph shoots up or down towards infinity. In general, to find vertical asymptotes, you are going to find out what value of x "fucks your equation up." by "fuck the equation up" we generally mean something like dividing by zero. For instance, if your equation is y = 1/x. We clearly see that x=0 would result in a divide by zero, thus there is a vertical asymptote at x=0. Or, if our graph was y = 1/(1-x^2), we see that both x=1 and x=-1 result in a divide by zero. Therefore, we have vertical asymptotes x=1 and x=-1. Its important to note that "x = NUMBER" is the equation for a line. Your answer MUST be in the form "x = crap" or else your teacher will be pissed at you. (some students just say "we see asymptotes at 1 and 7" which is incorrect. you say "we have asymptotes at x=1 and x=7"

4. horiz. asymp:
For these, you just see what happens when you let x go towards infinity AND negative infinity. So, if your equation was y=1/x. We see that plugging in infinity yields y=0 and plugging in negative infinity yields y=0 as well. Therefore, we would have a horizontal asymptote at y=0. (its just a coincidence in my example that you got the same thing by plugging in both positive and negative infinity. always check for BOTH)

5. relative extrema:
relative extrema refer to the minimum and maximum points (peaks and valleys) along the curve. relative means "all points that are peaks or valleys." if the problem asked for the "absolute extrema" it would want the coordinates of the highest peak and the lowest valley. Since it just asks for relative extrema, this is pretty easy.

So, the derivative of the function tells you whether the function is increasing or decreasing at any point. So if y = x^2, we'd know y' = 2x. Therefore, at x=1, the function is increasing at a rate of 2. Likewise, at -2, the function is decreasing at a rate of 4. Therefore, since we are looking for peaks and valleys, these are points where the function ISNT increasing or decreasing, its just flat. In other words, these are the points where the derivative is ZERO.

So, to find your relative min/max, you simply take the derivative and set it equal to zero. This will give you the x-coordinates of the min/max (you'll need to find the corresponding y-coordinates by plugging the x-coordinates into the original function).

You're not done yet though!! For each point, you need to know whether its a minimum or a maximum. To do this, you use the second derivative. The second derivative tells you whether or not the function is concaved up or down (concave up = valley, concave down = peak). So, simply take an x-coordinate that's a min/max point, plug it into the SECOND derivative, and look at whether your answer is positive or negative. If the answer is positive, your function is concave up, and you have a minimum at your x-coordinate. if the answer is negative, your function is concaved down, and you have a maximum at your x-coordinate.

6. points of inflection:
If you are skiing down a mountain, the inflection point would be the point at which you'd say "OH SHIT" because its the point at which you are going the fastest. Using the peaks and valleys idea, if you start at the top of a peak, and start rolling down the side, you'll keep gaining speed as long as the hill is getting steeper and steeper. You'll start slowing down once you approach the valley and your hill is getting less steep. In other words, you are going the FASTEST right at the moment when the concavity changes (from concave up to concave down).

Therefore, to find the inflection points, take the SECOND derivative, and then set it equal to zero. Solving for x yields the x-coordinates of your inflection points. Again, plug these x-values back into the original function to find the coordinates of the inflection point

DONE EZ
Whenever I encounter some little hitch, or some of my orbs get out of orbit, nothing pleases me so much as to make the crooked straight and crush down uneven places. www.day9.tv
Raithed
Profile Blog Joined May 2007
China7078 Posts
June 04 2008 19:44 GMT
#7
wow day, thanks lol.
Day[9]
Profile Blog Joined April 2003
United States7366 Posts
June 04 2008 19:48 GMT
#8
haven't done calc in a long time

nostalgia t.t
Whenever I encounter some little hitch, or some of my orbs get out of orbit, nothing pleases me so much as to make the crooked straight and crush down uneven places. www.day9.tv
gwho
Profile Blog Joined January 2008
United States632 Posts
June 04 2008 21:22 GMT
#9
good guide, day9. thorough and beyond~. rewriting the textbook lol.

gl on ur hw=)
knyttym
Profile Blog Joined December 2006
United States5797 Posts
June 04 2008 22:47 GMT
#10
On June 05 2008 04:38 Day[9] wrote:


3. vertical asymp:
A vertical asymptote is a point at which the graph shoots up or down towards infinity. In general, to find vertical asymptotes, you are going to find out what value of x "fucks your equation up." by "fuck the equation up" we generally mean something like dividing by zero. For instance, if your equation is y = 1/x. We clearly see that x=0 would result in a divide by zero, thus there is a vertical asymptote at x=0. Or, if our graph was y = 1/(1-x^2), we see that both x=1 and x=-1 result in a divide by zero. Therefore, we have vertical asymptotes x=1 and x=-1. Its important to note that "x = NUMBER" is the equation for a line. Your answer MUST be in the form "x = crap" or else your teacher will be pissed at you. (some students just say "we see asymptotes at 1 and 7" which is incorrect. you say "we have asymptotes at x=1 and x=7"

[


If math textbooks were written like this, it would be so much easier to learn
paper
Profile Blog Joined September 2004
13196 Posts
June 04 2008 23:21 GMT
#11
geeeeeeeeeeeeeeeeeeeeeeeeeeeeeeez raithed do you even read your damn textbook
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