`y=-3/(x+2)` Graph the function. State the domain and range.
To be able to graph the given function `y=-3/(x+2)` , we need to solve for the possible location of vertical asymptote.
Vertical asymptote exist at x=a that will satsify `D(x)=0` on a rational function `f(x)=(N(x))/(D(x))` .
To solve for the verical asymptote, we equate the expression at denominator side to 0 and solve for x.
A vertical asymptote exist along `x=-2` .
To solve for horizontal asymptote for a given function: `f(x) = (ax^n+...)/(bx^m+...)` , we follow the conditions:
when `n lt m` horizontal asymptote: `y=0`
`n=m ` horizontal asymptote: ` y =a/b`
`ngtm ` horizontal asymptote: `NONE`
The function `y=-3/(x+2)` is the same as `y=(-3x^0)/(x^1+2)` .
Then, `n=0` and `m=1` satisfies the condition: n<m.
Therefore, a horizontal asymptote exist at `y=0` (along x-axis).
To solve for possible y-intercept, we plug-in `x=0 ` and solve for `y ` .
`y=-3/2 or -1.5 `
Then, y-intercept is located at a point `(0,-1.5) ` .
To solve for possible x-intercept, we plug-in `y=0 ` and solve for x.
The x's get cancelled out. Thus, there is no x-intercept.
The y-intercept `(0,-1.5)` indicates that the graph is below the x-axis. Given that we can not cross the x-axis due to the horizontal asymptote, it follows that the graph approach the vertical asymptote in downward direction from right and upward direction from the left.
Solve for additional points as need to sketch the graph.
When `x=-5` , then `y =-3/(-5+2)=1` . point: `(-5,1)`
When `x=-3` , then `y =-3/(-3+2)=3` . point: `(-1,3)`
When `x=1` , then` y =-3/(1+2)=-1` . point: `(1,-1)`
Applying the listed properties of the function, we plot the graph as:
The domain of the function is based on the possible values of x.
Domain: `(-oo, -2)uu(-2,oo)`
`x=-2` excluded due to the vertical asymptote
The range of the function is based on the possible values of y.
`y=0` is excluded due to the horizontal asymptote.