Find the parabola of the form y=ax2+b which best fits the points (1,0), (2,2), (4,4) by minimizing the sum of squares, S, given by

Step-by-step explanation:

As the slope between (1,0) and (2,2) is 2 and the slope between (2,2) and (4,4) is 1, we can tell that the parabola must be downward opening and have a vertex in the first quadrant

Possibly best to use the vertex form to make 3 equations with 3 unknowns

y = a(x - h)² + k

0 = a(1 - h)² + k  (i)

2 = a(2- h)² + k  (ii)

4 = a(4 - h)² + k  (iii)

subtract i from ii

2 = a((2 - h)² - (1 - h)²)

2 = a(4 - 4h + h² - (1 - 2h + h²))

2 = a(3 - 2h)

a = 2/(3 - 2h)

subtract i  from iii

4 = a((4 - h)² - (1 - h)²)

4 = a(16 - 8h + h² - (1 - 2h + h²))

4 = a(15 - 6h)

substitute for a

4 = (2/(3 - 2h))(15 - 6h)

4(3 - 2h) = 2(15 - 6h)

12 - 8h = 30 - 12h

4h = 18

h = 9/2

a = 2/(3 - 2(9/2))

a = - 1/3

0 = -1/3(1 - 9/2)² + k

k = 49/12

y = (-1/3)(x - 9/2)² + 49/12

expand to get y = ax² + bx + c form

y = (-1/3)(x² - 9x + 81/4) + 49/12

y = (-1/3)x² + 3x - 27/4) + 49/12

y = (-1/3)x² + 3x - 8/3

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

~

answer: 12

step-by-step explanation:

because yes