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

1 Answer
May 13, 2018

" "
**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.