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

May 13, 2018

#### Answer:

$\text{ }$
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:

$\text{ }$
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

$\left(h , k\right)$ 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)

$\text{Range :} f \left(x\right) \ge 8$

Using Interval Notation : $\left[8 , \infty\right)$

color(green)("Step 2: "

Find Domain :

Domain of $f \left(x\right)$ 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:

Hope it helps.