How do you find the domain and range of #y=x^2+1#?

1 Answer
Nov 15, 2017

Answer:

Domain: #x in RR or x| (-oo,oo)#
Range : # y>= 1 or y| [1,oo)#.

Explanation:

#y=x^2+1 # , Domain : Possible input value of #x# is

any real value . Therefore Domain: #x in RR or x| (-oo,oo)#.

Range: #y=x^2+1 or y = (x-0)^2+1 # . Comparing with vertex

form of equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex

we find here #h=0 , k=1,a=1 :.# Vertex is at #(0,1)#

Since #a# is positive the parabola opens upward and

vertex is the minimum point #x=0, y=1#

So range is #y>=1 or y| [1,oo)#.

Domain: #x in RR or x| (-oo,oo)#

Range : # y>= 1 or y| [1,oo)#.

graph{x^2+1 [-10, 10, -5, 5]}