|x2+4|

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

(-∞,∞)

Set-Builder Notation:

{x|x∈ℝ}

Substitute the x value -2 into f(x)=|x2+4|. In this case, the point is (-2,8).

Replace the variable x with -2 in the expression.

f(-2)=|(-2)2+4|

Simplify the result.

Raise -2 to the power of 2.

f(-2)=|4+4|

Add 4 and 4.

f(-2)=|8|

The absolute value is the distance between a number and zero. The distance between 0 and 8 is 8.

f(-2)=8

The final answer is 8.

y=8

y=8

y=8

Substitute the x value -1 into f(x)=|x2+4|. In this case, the point is (-1,5).

Replace the variable x with -1 in the expression.

f(-1)=|(-1)2+4|

Simplify the result.

Raise -1 to the power of 2.

f(-1)=|1+4|

Add 1 and 4.

f(-1)=|5|

The absolute value is the distance between a number and zero. The distance between 0 and 5 is 5.

f(-1)=5

The final answer is 5.

y=5

y=5

y=5

Substitute the x value 0 into f(x)=|x2+4|. In this case, the point is (0,4).

Replace the variable x with 0 in the expression.

f(0)=|(0)2+4|

Simplify the result.

Raising 0 to any positive power yields 0.

f(0)=|0+4|

Add 0 and 4.

f(0)=|4|

The absolute value is the distance between a number and zero. The distance between 0 and 4 is 4.

f(0)=4

The final answer is 4.

y=4

y=4

y=4

Substitute the x value 1 into f(x)=|x2+4|. In this case, the point is (1,5).

Replace the variable x with 1 in the expression.

f(1)=|(1)2+4|

Simplify the result.

One to any power is one.

f(1)=|1+4|

Add 1 and 4.

f(1)=|5|

The absolute value is the distance between a number and zero. The distance between 0 and 5 is 5.

f(1)=5

The final answer is 5.

y=5

y=5

y=5

The absolute value can be graphed using the points around the vertex (-2,8),(-1,5),(0,4),(1,5)

xy-28-150415

xy-28-150415

Graph |x^2+4|