# Thread: sketch the graph of f(x)

1. ## sketch the graph of f(x)

determine the horizontal and vertical asymptotes of the function f(x) = 1/(4-x^2) and sketch the graph of f(x)...

my attempts to the question :

i rearrange the graph f(x) = 1/-(x^2-4) ==> 1/-((x-2)(x+2))

so it has two vertical assymptotes which are x = -2 and 2. is there any horizontal assymptote??? huhuhu

by using the conventional method to sketch the graph the graph of f(x) can be obtained by :
1. reflecting graph 1/x^2 by y-axis and shift the graph 4 unit to the right...

but when i sketch the graph of f(x) by using gc it looks like in the figure...

since in the exam we are not allow to bring gc, is there any other ways to sketch the graph??? or my way to treat this question is wrong??? PLZ HELP ME...

2. Originally Posted by bobey
determine the horizontal and vertical asymptotes of the function f(x) = 1/(4-x^2) and sketch the graph of f(x)...

my attempts to the question :

i rearrange the graph f(x) = 1/-(x^2-4) ==> 1/-((x-2)(x+2))

so it has two vertical assymptotes which are x = -2 and 2. is there any horizontal assymptote??? huhuhu

by using the conventional method to sketch the graph the graph of f(x) can be obtained by :
1. reflecting graph 1/x^2 by y-axis and shift the graph 4 unit to the right...

but when i sketch the graph of f(x) by using gc it looks like in the figure...

since in the exam we are not allow to bring gc, is there any other ways to sketch the graph??? or my way to treat this question is wrong??? PLZ HELP ME...
Sure,

$f(x)=\frac{1}{4-x^2}$

As x approaches infinity to the left or right, the denominator goes to $-\infty$

as there is no conceiveable difference between $4-\infty$ and $-\infty$

so the expression goes to zero.

Hence the graph approaches the x-axis as f(x) approaches zero

3. why suddenly in the graph got sharp turn point on the right and the left of the graph... i really could not understand this... can anyone explain to me with prove rather than giving explaination without proving on how to sketch the graph of f(x) from graph of 1/x^2

4. Here is more detail on the graph, using more values of x.
Your machine is not using very much of the domain to draw the graph.

5. Originally Posted by Archie Meade
Here is more detail on the graph, using more values of x.
Your machine is not using very much of the domain to draw the graph.

can you explain to me on how to get the graph of f(x) by transformation of graph 1/x^2??? since we are not allowed to bring gc in exam hall.. graph of 1/x^2 got two graph on the left and on the right and suddenly the graph above got three graph, i.e., on the left, right + AT THE CENTER??? hoe this happened? still blurrrrrrr

6. If you sketch a few solutions of these types of problems,
or just examine the graphs until you understand clearly what's happening on the graph,
you will master it.

When the denominator approaches zero, the graph is approaching a vertical asymptote.

Vertical asymptotes are located on the x that causes the denominator to be zero.

To find these, you factorise the denominator.
Each of the factors then show the x causing that factor to be zero.

$f(x)=\frac{1}{4-x^2}=\frac{1}{(2-x)(2+x)}$

Hence, there are vertical asymptotes at $x=2,\ x=-2$

To understand the shape of the graph, pick x "very near" x=-2 and 2.
If x is slightly less than -2, f(x) is negative and goes to negative infinity in this case.
If x is slightly greater than -2, f(x) is positive and the graph goes to infinity.

Do the same analysis at x=2.

The part in the middle comes down from infinity and reaches $\frac{1}{4}$ when x=0.

The part in the middle is always positive.

Finally, you must understand how to calculate the limit of f(x) as x approaches plus and minus infinity.

Take the function given and work with that rather than trying to compare with another function.

In this case, because of the 2 vertical asymptotes,
the graph is split into 3 regions,
one to the left of x=-2,
another to the right of x=2,
the other between x=-2 and x=2.