1. ## curve sketching

Been trying to get past this but it just keeps building on each other. It's for tomorrow:

$f(x)=\frac{x}{(x+1)(x+2)}$

a) Find any vertical and horizontal asys
b) where axis are crossed
c) show curve has no turning points
d) sketch curve, labeling it properly.
e) sketch f(-x) on the graph.

So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?

Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.

Curve's turning points. I want to see if differentiating $\frac{dy}{dx}$ should be enough, or do i need to go down to $\frac{d^2y}{dx^2}$ and make it =0. If the answer is imaginary it's proved, correct?

sketching curve should be done using the information gained from the previous questions and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.

f(-x) means a reflection in the x axis? Thanks for the posts

2. Originally Posted by Lonehwolf
Been trying to get past this but it just keeps building on each other. It's for tomorrow:

$f(x)=\frac{x}{(x+1)(x+2)}$

a) Find any vertical and horizontal asys
b) where axis are crossed
c) show curve has no turning points
d) sketch curve, labeling it properly.
e) sketch f(-x) on the graph.

So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?
that's right. so the vertical asymptotes are x = -1 and x = -2

for the horizontal asymptotes, find $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$. if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something

Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.
don't know what you mean. this is finding the x- (and y-) intercept

Curve's turning points. I want to see if differentiating $\frac{dy}{dx}$ should be enough, or do i need to go down to $\frac{d^2y}{dx^2}$ and make it =0. If the answer is imaginary it's proved, correct?
for turning points, you only need the first derivative, and set it to zero.

sketching curve should be done using the information gained from the previous questions

and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.
huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.

f(-x) means a reflection in the x axis? Thanks for the posts
nope, you reflect in the y-axis

3. Originally Posted by Jhevon
that's right. so the vertical asymptotes are x = -1 and x = -2

for the horizontal asymptotes, find $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$. if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something

don't know what you mean. this is finding the x- (and y-) intercept

for turning points, you only need the first derivative, and set it to zero.

huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.

nope, you reflect in the y-axis
Can't work the horizontal asymptotes either way. Not sure how you work with $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$.

4. Originally Posted by Lonehwolf
Can't work the horizontal asymptotes either way. Not sure how you work with $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$.
do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote

5. Originally Posted by Jhevon
do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote
Uhh, I do know about limits, but i can't recall those wicked symbols :

How are they used, I must have been sick during that lesson!

6. Originally Posted by Lonehwolf
Uhh, I do know about limits, but i can't recall those wicked symbols :

How are they used, I must have been sick during that lesson!
$\lim_{x \to \infty} f(x)$ reads "the limit as $x$ goes (or tends) to $\infty$ of $f(x)$". you pretty much evaluate what will happen if $x$ gets huge. if $x$ gets really big, beyond measure, the function here, will get close to zero.

a similar interpretation holds for $\lim_{x \to - \infty}f(x)$

the same symbol is used in the definition of the derivative, so you must have been sick during several lessons

7. Originally Posted by Jhevon
$\lim_{x \to \infty} f(x)$ reads "the limit as $x$ goes (or tends) to $\infty$ of $f(x)$". you pretty much evaluate what will happen if $x$ gets huge. if $x$ gets really big, beyond measure, the function here, will get close to zero.

a similar interpretation holds for $\lim_{x \to - \infty}f(x)$

the same symbol is used in the definition of the derivative, so you must have been sick during several lessons
Thanks for the explanation, I believe this issue is solved, now i gotta start a clean sheet and burn that whole mess of a paper i had.