What is the vertex of this parabola?
y-5 = x^2?

Work Shown:

y-5 = x^2

y = x^2 + 5 ... add 5 to both sides

y = 1x^2 + 5

y = 1(x-0)^2 + 5 ... x replaced with (x-0)

The equation is now in the form y = a(x-h)^2 + k

We have a = 1, h = 0, k = 5.

The vertex is (h,k) = (0,5)

The value of 'a' determines how vertically stretched or compressed the parabola is. Also, if a > 0, then the parabola opens upward; or if a < 0, then the parabola opens downward.

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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