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Analyzing a Function

Date: 05/19/97 at 14:08:48
From: Jessica
Subject: Graphing

This is a homework question that I cannot figure out. We have to do a 
complete graphical analysis, stating the domain, intercepts, 
asymptotes, critical points and points of inflection.  We also have to 
graph the function:

  y = 3x^4-4x^3-12x^2

I can't find any x-intercepts, even using Newton's method.

Any help that you can give me would be appreciated.

Date: 05/25/97 at 11:06:50
From: Doctor Rob
Subject: Re: Graphing

Since any real-number value of x gives a corresponding y-value, the
domain is the real numbers.

y-intercept:  (0,1)
x-intercept:  e^x is always greater than 2*x, so there is no 
x-intercept.  (You have to supply a proof of this, of course!  HINT: 
It can involve the parts of the problem where you find critical points 
and inflection points.)

Critical points are points where dy/dx changes sign. To find them,
compute dy/dx, set the result equal to zero, and solve for x. Then 
among the resulting values of x, you must find those where dy/dx is 
positive on one side and negative on the other. If it changes from 
plus to minus going from smaller values of x to larger ones, then you 
have a local maximum. If minus to plus, a local minimum.

Inflection points are points where d^2y/dx^2 changes sign. This is
similar to the previous paragraph. You compute the second derivative, 
set the result equal to zero, and solve for x. Among the resulting 
values of x, you must find those where it is positive on one side and 
negative on the other. These are the inflection points.

Asymptotes are lines to which the function is tangent at infinity. 
There are several ways of finding them. The usual thing to do is look
for horizontal and vertical asymptotes. Vertical asymptotes occur at
values of x that are boundary points of the domain, where y heads 
toward infinity as x heads toward a. Horizontal asymptotes occur at 
the limiting y-values as x heads toward positive or negative infinity.
Here it helps to know the asymptotes of some simple functions.  
y = b + 1/(x-a) has a vertical asymptote at x = a, and a horizontal 
one at y = b.  y = a^x, a > 1, has a horizontal asymptote at y = 0.  
y = log_a(x) has a vertical asymptote at x = 0. y = arctan(x) has 
vertical asymptotes at x = (2*k+1)*Pi/2, for every integer value of k, 
and so on.

Another approach to asymptotes is to replace x by 1/u and y by 1/v, to
get a new curve F(u,v) = 0. What happens to this new curve at (0,0)
determines the asymptotes of the old curve. In particular, the 
tangents to F = 0 at (0,0) have slopes that are the reciprocals of the 
slopes of the asymptotes of the original curve. You can get these 
slopes by implicitly differentiating F(u,v) = 0 and solving for dv/du 
at u = 0 and v = 0.

To graph the curve, pick a few values of x and compute the 
corresponding values of y. Plot those points. Knowing the other parts 
above will help you decide how to connect the dots. It will also help 
you to know where to take more points (near vertical asymptotes, for 
example), and where to take fewer.

I hope this helps!

-Doctor Rob,  The Math Forum
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Associated Topics:
High School Calculus

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