Sketch graph of the function f(x) = |x+2|.

The absolute value can be graphed using the points around the vertex (-2, 0), (-4, 2), (-3, 1), (-1, 1), (0, 2).

Step-by-step explanation:

The find the x coordinate of the vertex, set the inside of the absolute value x + 2 equal to 0. In this case, x + 2 = 0.

x + 2 = 0

Subtract 2 from both sides of the equation.

x = -2

Replace the variable x with -2 in the expression.

y = |(-2) + 2|

The absolute value vertex is (-2, 0).

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.

For each x value, there is on y value. Select few z 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 -4 into f(x) = |x+2|. In this case, the point is (-4,2).

y=2

Substitute the x value -3 into f(x) = |x+2|. In this case, the point is (-3,1).

y=1

Substitute the x value 0 into f(x) = |x+2|. In this case, the point is (0,2).

y=2

You can find the graph in the attachment.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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