Question

Question #dea4c

Answer 1

The vertical asymptotes are: `x=-2` and `x=-5`

Explanation:

You need to factor the denominator and find the zeros, and that will give you the vertical asymptotes.

so, `y=(x-2)/((x+2)(x+5))`

equate denominators to zero, and you will get the zeros, which are the vertical asymptotes:

`x+2=0`
`x=-2`

and

`x+5=0`
`x=-5`

As well, note that the zeros are not holes due to the fact that they do not cancel out!

Hope this helps!

Answer 2

`"vertical asymptotes at "x=-5" and "x=-2`
`"horizontal asymptote at "y=0`

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve "x^2+7x+10=0rArr(x+5)(x+2)=0`

`rArrx=-5" and "x=-2" are the asymptotes"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),ytoc" ( a constant)"`

`"divide terms on numerator/denominator by the highest"`
`"power of x, that is "x^2`

`y=(x/x^2-2/x^2)/(x^2/x^2+(7x)/x^2+10/x^2)=(1/x-2/x^2)/(1+7/x+10/x^2)`

as `xto+-oo,yto(0-0)/(1+0+0)`

`rArry=0" is the asymptote"`
graph{(x-2)/(x^2+7x+10) [-10, 10, -5, 5]}