# State the asymptotes, holes, roots and intercepts for 2x^2+10x+12/x^2-9Include Vertical asymptote, horizontal asymptote, roots, x-int and y-int with end behaviours and describe how the graph will...

State the asymptotes, holes, roots and intercepts for 2x^2+10x+12/x^2-9

Include Vertical asymptote, horizontal asymptote, roots, x-int and y-int with end behaviours and describe how the graph will look like

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### 2 Answers

f(x)=(2x^2+10x+12)/(x^2-9)

We can factor to get

f(x)=(2(x^2+5x+6))/((x+3)(x-3))=(2(x+3)(x+2))/((x+3)(x-3))

We can simplify with the idea that x!=-3 or x!=3

f(x)=(2(x+2))/(x-3)

Now we have vertical asymptotes at x=3, and a hole at x=-3.

Since deg(2(x+2)) = deg(x-3) and the leading coefficients are 2 and 1 we get a horizontal asymptote at y=2.

The x-intercept is when 2(x+2) = 0 or x=-2.

The y-intercept is when y=(2(0+2))/(0-3)=-4/3

The end behavior is y=2.

The below graph is how we would plot this funcion:

Note the hole at x=-3, it is not an asymptote.

`f(x)=[2x^2+10x+12]/[x^2-9]=>`

`f(x)=[2(x^2+5x+6)]/[(x-3)(x+3)]=>`

`f(x)=[2(x+2)(x+3)]/[(x-3)(x+3)]`

1) Since x=3, and x=-3 get the denominator to be zero than they are your certical asymptote.

2) Since the numerator and denominator are of the same degree we have a horizontal asymptote, y=2.

3) No oblique asymptote.

4) Roots

`f(x)=0=>2(x+2)(x+3)=0=>x+2=0=>x=-2`

`x+3!=0`

Hence x-intercept (-2,0)

5) y-intercept

`x=0=>y=[2*(2)*(3)]/[(0+3)(0-3)]=>y=12/-9=-4/3`