This is a parabola word problem from pre calculus?

y = (- 16/2025)x² + 9/5 x + 1.7

y = - 0.0079x² + 1.8x + 1.7

a.) Initial height is given by the constant in the equation, or 1.7 ft.

b.) For maximum height, convert to vertex form, y = a(x - h)² + k,

where (h, k) is the vertex and k is the maximum height:

y = (- 0.0079x² + 1.8x) + 1.7

y = - 0.0079(x² - 227.85x) + 1.7

y = - 0.0079(x² - 227.85x + 12978.7) + 1.7 - [- 0.0079(12978.7)]

y = - 0.0079(x - 113.93)² + 1.7 - (- 102.53)

y = - 0.0079(x - 113.93)² + 1.7 + 102.53

y = - 0.0079(x - 113.93)² + 104.23

Vertex (113.93, 104.23), so

Max. Ht. = 104.23 ft.

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c. ) Max. Distance: y = 0:

- 0.0079x² + 1.8x + 1.7 = 0

- 0.0079x² + 1.8x = - 1.7

- 0.0079(x² - 227.85x) = - 1.7

x² - 227.85x = - 1.7 / - 0.0079

x² - 227.85x = 215.2

x² - 227.85x + 12978.7 = 12978.7 + 215.2

(x - 113.93)² = 13193.9

x - 113.93 = √13193.9

x - 113.93 = ± 114.87

x = 113.93 ± 114.87

If x = 113.93 + 114.87,

x = 228.79

If x = 113.93 - 114.87,

x = - 0.94

Feasible Domain: x > 0

Since x = - 0.94 is not in the feasible domain,

x = 228.79

Max. Distance = 228.79 ft.

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i) 1.7 ft

ii) dy/dx = -32x/2025 + 9/5 = 0 for maximum

x = 113.9

y = 104.2 ft

iii) 0 = (-16/2025)x^2 + (9/5)x + 1.7

Use the quadratic formula

x = 228.8 ft