Question

How do you identify the zeroes and asymptotesof `f(x) = (4x^2-16)/(x^2-9)`?

Answer 1

Let me split this up

Explanation:

To find zeros, it means to find x-intercepts, i.e. where `f(x)=0`
In this case, when `f(x)=0`,
`(4x^2-16)/(x^2-9)=0`
Assuming `x^2-9` does not equal zero, we multiply the whole equation by `x^2-9`, giving `4x^2-16=0(x^2-9)`
`therefore 4x^2-16=0`
`x^2=16/4`
`x=+-sqrt4=+-2`
Verifying that `x=+-2` does not result in `x^2-9` to equal 0, (it equals `-5`)
Therefore, `x=+-2` are zeroes for `f(x)`

To find asymptotes, we have to find the x and y values as y and x tend to infinity. Let me split this up into horizontal and vertical asymptotes:

Horizontal asymptote is the y-value as `x->+-oo`, i.e.
`x->+-oo (4x^2-16)/(x^2-9)`
Dividing the numerator and denominator by the variable with the highest power, i.e. `x^2`,
`x->+-oo (4x^2-16)/(x^2-9)=x->+-oo (4-16/x^2)/(1-9/x^2)`
As the variable in the fraction is a squared variable, it means that `x->-oo (4-16/x^2)/(1-9/x^2)=x->+oo (4-16/x^2)/(1-9/x^2)` as the negative sign is removed through the square.
`thereforex->+-oo (4-16/x^2)/(1-9/x^2)=x->oo (4-16/x^2)/(1-9/x^2)=x->oo(4-0)/(1-0)=4`

Therefore, the horizontal asymptote is `y=4`

Vertical asymptotes are x-values that cause the function to carry a not well defined value or indeterminate form, such as `"something"/0`. In other words, to find the vertical asymptote means to solve for `x` when the denominator equals 0, i.e.
`x^2-9=0`
`x^2=9`
`x=+-3`
This means that at `x=+3` and `x=-3`, `f(x)` is undefined, hence the vertical asymptotes for `f(x)` are `x=+-3`