Can some one explain to me how to graph the function of f(x)= 4/x^2+1. I'm not sure where to start.?

Question

By looking at it im pretty sure that the graph is going to be a parabola because of the x^2. Also that it is shifted one unit to the left because of the plus 1. I do not know what to do with the 4 in the numerator.
Is this a composition of transformations?
Do I have to simplify it into two functions and how do I do it?
What do I do with that 4?!!
Thanks,
Super Confused....

Answer 1

it's not going to be a parabola because the x^2 is on the bottom, for it to be a parabola the x^2 would have to be in the numerator

Answer 2

Consider y = 4/x^2+1

let x = 0, then y = 4/1 4
when x = 1, y = 4/1+1 = 4/2 = 2
when x = 2, y = 4/2^2 + 1 = 4/4+1 = 4/5
when x = 3, y = 4/3^2+1 = 4/9+1 = 4/10 = 2/5
when x = 4, y = 4/4^2+1 = 4/16+1 = 4/17

Plot these vales for x & y on a graph and you get a graph.

Answer 3

When I can't visualize something I use a graphing applet... the resulting graph is given in the second link.

Because you have x^2 in the denominator, it is not a parabola.

One thing to try is to plug in some values. A value of x = 0 results in a value of 4 / 1 = 4. So it crosses the y-axis at (0,4).
Also, because of the x^2, it will be symmetric around the y-axis.

There is no way to make y be zero. Nor can you make it negative. So y will all be above the x axis.

Basically, just keep filling in values of x (1/2, 1, 2, etc.) and connect the dots. You'll get the same values for (-1/2, -1, -2) so just mirror it on the left. The result is the graph below (2nd link)

Answer 4

something is a function if f(x) is uniquely defined for all x interior the area. Uniquely defined skill f(x) can't equivalent 2 values on a similar time - ie, any vertical line cuts the graph at maximum as quickly as. It extremely relies upon on the question - in case you basically propose from a graph, merely look up the place the graph hits the cost for f(x), and study off x off the axis. yet for algebraic strategies, it extremely relies upon.

Answer 5

Just make a table with various values of x in the left column and work out the values of y from the formula you have in the right column. This will give you a set of points to graph.

BTW, the x^2 is in the denominator, so it won't be a parabola. And the +1 will cause the whole curve to move upward 1.0 (no effect on the x direction)

Answer 6

f(x)= 4/x^2+1

Undefined when x = 0 (therefore a vertical asymptote)
As x → ±∞, 4/x^2 → 0 so f(x) → 1 both ways ie asymptotic to y = 1

Since x^2 >0 for all x > 0, f(x) >1 for all x

So it 'looks like a hyperbola with both parts above y = 1 and asymptotic to y = 1 and x = 0