Question

Find the x-intercepts of the polynomial f(x) = -x^2+16

Answer 1

We want to find the x-intercepts of the polynomial

`f(x)=\text{ -x}^2+16`

this means, we want to find where this functon crosses the x axis. Recall that the x axis is the line y=0. So we want to solve this equation

`0=\text{ -x}^2+16`

if we multiply both sides by -1, we get

`0=x^2\text{ -16}`

on the right, we have a difference of squares, so we can factor it out as

`0=(x+4)\cdot(x\text{ -4\rparen}`

now, as this is a product of numbers, this means that each of the number could be 0. This means we have two different equations, which are

`x+4=0`

and

`x\text{ -4=0}`

on the first one, if we subtract 4 on both sides, we get

`x=\text{ -4}`

and on the second one, if we add 4 on both sides, we get

`x=4`

so the x-intercepts of the polynomial are x=4 and x= -4