# sketch the graph of the curve (x-1)/(x^2-4)also find symmetricity intercepts asymptotes region of increasing and decreasing critical points(points of extrema) points of inflexion regions of concavity

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### 1 Answer

Sketch the graph of `f(x)=(x-1)/(x^2-4)` :

First factor numerator and denominator:

`f(x)=(x-1)/((x+2)(x-2))`

(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

`f'(x)=((x^2-4)(1)-[(x-1)(2x)])/((x^2-4)^2)`

`=-(x^2-2x+4)/((x^2-4)^2)`

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

`f''(x)=((x^2-4)^2(-2x+2)-[(-x^2+2x-4)(2)(x^2-4)(2x)])/((x^2-4)^4)`

`=(2(x^3-3x^2+12x-4))/((x^2-4)^3)`

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)`

The graph: