# Graph |x^2+4| |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}
For each x value, there is one y value. Select few x values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.
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|
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
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|
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
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|
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
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|
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
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|     