# Curve Sketching

1. Dec 23, 2004

### majinknight

Hi i was wondering if somone could check to see if i did this question correctly.
The question is: For the function y=3x^2 -1/x^3
a)Determine the x-intercept(s) and y-intercept(s)
b)Determine any asymptotes
c)Find the intervals of increase and decrease
d)Final all maximums and minimums
e)Find the interval(s) where the graph is concave upward and concave downward
f)Find any points of inflection

a) when x=0, y=-1/0, so the y-intercept is not in the real number plane.
when y=0, x^2=1/3, so the x-intercepts are positive 1/square root of 3 and negitive 1/square root 3.

b)Verticle: x^3=0
x=0, there fore a verticle asymptote at x=0.

limit as x->0 from the left -1/small negitive number= positive infinity(shoots upward).
limit as x->0 from the right -1/small positive number=negitive infinity (shoots downward).

Horizontal: lim as x->infinity from the right 3x^2/x^3 =lim as x-> infinity from the right 3/x =0.
lim as x->infinity from the left =3/x =0
Therefore there is a horizontal asymptote at y=0.
For very large positive values of x the function approaches the horizontal asymptote from above. For very "large" negitive values of x the fucntion approaches the horizontal asymptote from below.

Slant: There is no slant asymptote.

c)d) dy/dx= 6(x^3)- (3x^2 -1)(3x^2) / x^6
=-3x^4 + 3x^2 / x^6
=-3(x^2 -1) / x^4
0=x^2 - 1
x= positive and negitive one

So when i make my chart i get that a local minimum occurs at (-1, -2) and a local maximum at (1,2). The function is increasing when -1<x<1. The function is decreasing when x<-1 and x>1.

e)f) second derivitive= (-12x^3 + 6x)(x^6) - (-3x^4 +3x^2)(6x^5) / x^12
=6x^9 - 12x^7 / x^12
=6(x^2 - 2) / x^5
0=x^2 - 2
x= positive and negitive square root of 2

So when i do my chart i get that the function is concave down when x<- square root 2 or when negitive square root 2<x< square root 2. The function is concave up when x> square root two. A point of inflection is at x=square root two and y= 5/ 2 square root 2.

I hope that was not too confusing the way i put it, and if anyone can see a mistake i wouldnt very much appreciate it.

2. Dec 24, 2004

### HallsofIvy

Staff Emeritus
Looks pretty good to me!

I assume that you MEAN y= (3x2-1)/x3 and NOT y= (3x2)- (1/x3).

Yes, because 1/x3 is not defined at x= 0, neither is y: the graph has no y-intercept but does have a vertical asymptote at x= 0.

y= 0= (3x2-1)/x3 is the same as 3x2-1= 0 so
x= +/- &radic;(1/3) are the two x-intercepts.

Since the denominator has a higher degree than the numerator, there is a horizontal asymptote at y= 3 and no slant asymptote.

y'= ((6x)(x3)- (3x2-1)(3x2))/x6
= (6x4- 9x4- 3x2)/x6

which will be positive as long as -3x4- 3x2= -3x2(x2- 1) is positive which is the same as x2- 1 negative. Yes, that will be 0 at x= 1 or -1, It will be positive (and so y' is negative) for x< -1, y' will be positive for -1< x< 0. It stays positive for 0< x< 1, and is negative for x> 1.

That is: y is decreasing for x< -1, has a local minimum at x= -1 (when y= -2). increasing from -2 to infinity for -1< x< 0, increasing from -infinity to 2 for 0< x< 1, has a local maximum at x= 1 (when y= -2) and then decreases for x> 1.

The second derivative is just as you say. It is 0 at x= +/- &radic;(2), is negative for x< -&radic;(2), positive for -&radic;(2)< x< 0, negative for 0< x< &radic;(2), positive for x> &radic;(2).
That is: there are inflection points at +/- &radic;(2), the graph is convex downward for x< -&radic;(2), convex upward for -&radic;(2)< x< 0, convex downward for 0< x< &radic;(2), and convex upward for x> &radic;(2).

3. Dec 28, 2004

### majinknight

Alright Thanks!