Question

What is the zero of the function

Answer 1

`x=6` (or 6)

Explanation:

Set f(x)=0:

`f(x)=(x^2+x-42)/(x^2+3x-28)`

`0=(x^2+x-42)/(x^2+3x-28)`

`0=x^2+x-42`

`0=(x+7)(x-6)`

`x=-7` or
`x=6`

Now check both solutions:

`f(-7)=((-7)^2+(-7)-42)/((-7)^2+3(-7)-28)`

`f(-7)=(49-7-42)/(49-21-28)`

`f(-7)=(42-42)/(28-28)`

`f(-7)=(0)/(0)`

Therefore,
`x\\eq-7` because there is a hole at
`(-7,0)`. The denominator, in addition, can never be 0, so the function is undefined for
`x=-7`.

`f(6)=((6)^2+(6)-42)/((6)^2+3(6)-28)`

`f(6)=(36+6-42)/(36+18-28)`

`f(6)=(42-42)/(54-28)`

`f(6)=(0)/(24)`

`f(6)=0`

Since
`f(6)=0`, then
`x=6` is the only zero of the function.