Question

How do you find `h(-3)` given `h(a)=a^2-1+a`?

Answer 1

`h(-3)=5`

Explanation:

Substitute `-3` everywhere you see `a`: `h(a)<=>h(-3)`

`h(-3)=(-3)^2-1+(-3)`

`h(-3)=9-1-3`

Use PEMDAS to solve...

`h(-3)=5`