Question

How do you find the domain and range of `F(x)= (x-3)^2 +2`?

Answer 1

Domain: `{x|x in RR}`
Range: `{y|y >2, y in RR}`

Explanation:

'A quadratic function has a domain of all the real numbers.

The range depends on the direction of opening and the y-coordinate of the vertex. Since `a` in `y = a(x - p)^2 + q` is positive, `f(x)` opens up. Therefore, the vertex will be an absolute minimum. The vertex is given by `(p, q)` in `y = a(x - p)^2 + q`, so the y-coordinate of the vertex is `y = 2`. The range of `f(x)` is `[2, + oo)`.

Hopefully this helps!