Sketch the graph of `f(x)=(x-1)/(x^2-4)` :
First factor numerator and denominator:
(a) The y-intercept occurs when x=0; the y-intercept is `f(0)=(-4)/(-4)=1/4`
(b) The x-intercept(s) occur whebn y=0;
`(x-1)/((x+2)(x-2))=0` Note that a rational function, like a fraction, is zero if and only if the numerator is zero when the denominator is nonzero. Here the numerator is zero when x=1, and the denominator is nonzero at x=1.
The x-intercept is 1
(c) The graph of f(x) is symmetric with respect to the y-axis if f(-x)=f(x)
`f(-x)=((-x)-1)/((-x)^2-4)=(-x-1)/(x^2-4) != f(x)`
This is a function os the graph is not symmetric about the x-axis.
The graph of f(x) is symmetric about the origin if -f(-x)=f(x)
`-f(-x)=-((-x)-1)/((-x)^2-4)=-(-x-1)/(x^2-4)=(x+1)/(x^2-4)!=f(x)` so the graph is not symmetric about the origin. ** We knew it was not since the only zero was 1**
(d) There are vertical asymptotes at x=2 and x=-2 since the denominator is zero and the numerator nonzero at these x values.
(e) The horizontal asymptote is y=0. Depending on the class you are in you can show that the limit as x increases or decreases without bound is zero, or in a typical college algebra/precalculus class you will note that the degree of the numerator is less than the degree of the denominator.
Since this is a calculus class, show `klim_(x->+-oo)f(x)=0` by dividing numerator and denominator by `x^2` .
(f) Using the quotient rule we get
All extrema occur at critical points; i.e. when `f'(x)=0` or `f'(x)` fails to exist. Since the solution to `f'(x)=0` is nonreal, there are no critical points, thus no extrema.
(g) Using the quotient rule again we get
Using a graphing utility we find that `f''(x)=0==>x~~.362166` ; the second derivative changes sign at this point so there is a change of concavity there.
Checking intervals we find f(x) is concave down on `(-oo,-2)` , concave up on `(-2,0.362166)` , concave down on `(0.362166,2)` , and concave up on `(2,oo)`