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

1 Answer
Mar 24, 2017

Answer:

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!