Question

What are he intercepts of `y=(x+1)^2-2`?

Answer 1

The `x`-intercepts are at `(sqrt2-1)` and `(-sqrt2-1)` and the `y`-intercept is at `(0, -1)`.

Explanation:

To find the `x`-intercept(s), plug in `0` for `y` and solve for `x`.

`0 = (x+1)^2 - 2`

Add `color(blue)2` to both sides:
`2 = (x+1)^2`

Square root both sides:
`+-sqrt2 = x+1`

Subtract `color(blue)1` from both sides:
`+-sqrt2 - 1 = x`

Therefore, the `x`-intercepts are at `(sqrt2-1)` and `(-sqrt2-1)`.

To find the `y-`intercept, plug in `0` for `x` and solve for `y`:
`y = (0+1)^2 - 2`

Simplify:
`y = 1^2 - 2`

`y = 1 - 2`

`y = -1`

Therefore, the `y`-intercept is at `(0, -1)`.

Hope this helps!