Answer:
For the equation
[tex]y=\frac{x-6}{x^2+5x+6}[/tex]
There are vertical asymptotes at x=-2 and x=-3
There are no holes
Step-by-step explanation:
The equation for this graph can be factored in its denominator to get
[tex]y=\frac{x-6}{(x+2)(x+3)}[/tex]
This means that there are VA's at x=-3 and x=-2
(There are no holes as there is not a factor that cancels out.)
A function assigns the values. The asymptotes and the holes for the graph lie at x=-2 and x=-3.
A function assigns the value of each element of one set to the other specific element of another set.
The vertical asymptotes and the holes for the graph can be found by factorizing the denominator of the given function. Therefore, the factorisation of the denominator is,
x² + 5x + 6 =0
x² + 2x + 3x +6 =0
x(x+2)+3(x+2)=0
(x+2)(x+3)=0
x = -2, -3
Hence, the asymptotes and the holes for the graph lie at x=-2 and x=-3.
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