Try to draw the graph of the given function with some ordered pairs

\(h(x) = |x + 2|\)

x = -7, h(x) = 5
x = -6, h(x) = 4
x = -5, h(x) = 3
x = -4, h(x) = 2
x = -3, h(x) = 1
x = -2, h(x) = 0
x = -1, h(x) = 1
x = 0, h(x) = 2
x = 1, h(x) = 3
x = 2, h(x) = 4
x = 3, h(x) = 5

Let us check the option choices;

(A) h(x) = x + 2 when x ≥ 0 and x - 2 when x < 0
if we plug x = -1, h(x) must be -3
But we clearly see h(x) = 1

(B) h(x) = x + 2 when x ≥ 2 and -x + 2 when x < 2
if we plug x = 1, h(x) must be 1
But we clearly see h(x) = 3

(C) h(x) = x + 2 when x ≥ -2 and -x - 2 when x < -2
if we plug x = -3, h(x) must be 1
we clearly see h(x) = 1

Check 1 or 2 more values (distant)

if we plug x = -5, h(x) must be 3
we clearly see h(x) = 3

(D) h(x) = x + 2 when x ≥ -2 and x - 2 when x < 2
if we plug x = 1, h(x) must be -1
But we clearly see h(x) = 3

(E) h(x) = x + 2 when x ≥ 2 and -x + 2 when x < 0
if we plug x = -1, h(x) must be 3
But we clearly see h(x) = 1

Hence, option C