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

1 Answer
Noah G
Mar 24, 2017

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!

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