How do you find the domain and range of #f(x) = (x + 5)^2 + 8#?

1 Answer
May 13, 2018

Answer:

#" "#
Domain: #color(blue)([-oo < x < oo ]#

Using Interval Notation: #color(blue)((-oo,oo)#

Range: #color(blue)([f(x)>=8 ]#

Using Interval Notation: #color(blue)([8,oo)#

Explanation:

#" "#
Given: #color(red)(y=f(x)=(x+5)^2+8#

The Vertex Form of a quadratic function is:

#color(green)(y=f(x)=a(x-h)^2+k#, where

#(h,k)# is the Vertex.

Note: #color(blue)(a=1, h=-5, k=8#

#color(green)("Step 1: "#

Vertex is at : #color(blue)((-5,8)#

Note :

If #a>0#, then the Vertex is a Minimum Value.

Since, #a=1#, the Vertex is #color(blue)("Minimum at "(-5,8)#

#"Range :"f(x)>= 8#

Using Interval Notation : #[8, oo)#

#color(green)("Step 2: "#

Find Domain :

Domain of #f(x)# is the set of all input values for which the given function is real and well-defined.

#color(blue)(f(x) = (x+5)^2+8# does not have any undefined points.

The function does not have any domain constraints.

Therefore domain is given by

#color(blue)(-oo < x < oo#

Using Interval Notation :

#color(blue)((-oo , oo)#

Hence, the required solutions are:
Domain: #color(blue)([-oo < x < oo ]#

Using Interval Notation: #color(blue)((-oo,oo)#

Range: #color(blue)([f(x)>=8 ]#

Using Interval Notation: #color(blue)([8,oo)#

#color(green)("Step 3: "#

Draw the graph of #color(red)(y=f(x)=(x+5)^2+8# to verify the solutions:

enter image source here

Hope it helps.