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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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.
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calendarEducator since 2012
write301 answers
starTop subject is Math
`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`