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The domain contains time values for when the diver is above the water.
- 4.9t^2+3t+10=0
Let's use the Quadratic Formula.
Substitute values
Calculate power
Multiply
a(- b)=- a * b
a-(- b)=a+b
Add terms
We can use a calculator to find approximate values of these solutions. -3+sqrt(205)/-9.8&≈ -1.15 -3-sqrt(205)/-9.8&≈ 1.77 Since the solution we are looking for represents time, we need the positive solution. Using the lower and upper bounds we found, we can write the domain for which the function makes sense. Domain: 0≤ t≤ 1.77
The range contains height values.
2 &Quadratic function: &&f(x)= ax^2+ bx+ c &Axis of symmetry:&&x=-b/2 a &Vertex:&&(-b/2 a,f(-b/2 a)) The coefficients are the ones we used above. a= - 4.9 b= 3 c= 10 We can use these values to find the first coordinate of the vertex.
Substitute values
a(- b)=- a * b
- a/- b= a/b
a/b=a * 10/b * 10
a/b=.a /2./.b /2.
This means that the diver reaches the maximum after 1549 seconds. To find the maximum height, we can substitute this value in the expression for height.
The maximum height the diver reaches is approximately 10.46 meters above the pool. Using the lower and upper bounds we found, we can write the range for which the function makes sense. Range: 0≤ t≤ 10.46