Question

How do you find the important points to graph `f(x)=1/(x-3)`?

Answer 1

Domain , Range , Standard form of the equation, Non- defined points, etc

Explanation:

`f(x) = 1/(x-3)`

1. At first, we have to find the Domain of the function. For this, the denominator of the function should not become zero.
   When the denominator of the function approaches to the 0, the function approaches to the infinity.
   `rArr x-3!= 0`
   `rArr x!= 3`
   So, `D_f: x in R-{3}`
   Here, `x rarr 3 rArr f(x) rarr oo`
2. Then, we should check the Range of the function.
   `y = 1/(x-3)`
   `yx-3y=1`
   `yx=3y+1`
   `x= (3y+1)/y`
   So, `R_f : y in R- {0}`
   Here, `yrarr0 rArr xrarr oo`
3. Then, we should check for the Standard form of the equation.
   `y = 1/(x-3)`
   `xy=3y+1`
   It resembles with `xy=c` which is the equation for the Rectangular Hyperbola.
   So, it is also a Rectangular Hyperbola.
4. Draw the graph excluding the non-defined point in Domain & Range.
   graph{1/(x-3) [-7.02, 7.025, -3.51, 3.51]}

Answer 2

See explanations below

Explanation:

`f(x)=1/(x-3)`

As you cannot divide by `0`, `x!=3`

There is a verical asymptote, `x=3`

When `x=0`, `=>`, `y=-1/3`, the y-intercept is `(0,-1/3)`

We calculate the limits

`lim_(x->-oo)f(x)=0^-`

`lim_(x->+oo)f(x)=0^+`

There is a horizontal asymptote, `y=0`

`lim_(x->3^-)f(x)=-oo`

`lim_(x->3^+)f(x)=+oo`

We calculate the first derivative

`f'(x)=-1/(x-3)^2`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `3` `color(white)(aaaaaaaaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `↘` `color(white)(aaaa)` `||` `color(white)(aaaaaa)` `↘`

This is the general form of the curve.

graph{(y-1/(x-3))(y)(y-1000(x-3))=0 [-5.39, 10.42, -4.916, 2.986]}