To create a quadratic function, we need to find the vertex (the maximum point) and two other points on the path. Let's choose the vertex to be (5, 95) since this is halfway between the starting and ending point and within our maximum height range. The other two points will be the initial point (0, 0) and the point when the object returns to the ground, let's say (10, 0).
A quadratic function has the form y = a(x-h)^2 + k, where (h, k) is the vertex. Thus, the equation will be y = a(x - 5)^2 + 95.
Now we need to find the value of 'a'. We can use one of the other two points for this, let's use (0, 0).
0 = a(0 - 5)^2 + 95
-95 = 25a
a = -3.8
So our equation becomes y = -3.8(x - 5)^2 + 95. Since this quadratic goes downward, we know the maximum height occurs at the vertex, which we chose to be (5, 95). Now we just plug in the initial point and the point when the object returns to the ground to make sure our equation works for both.
Initial height:
y = -3.8(0 - 5)^2 + 95 = 0
Second zero:
y = -3.8(10 - 5)^2 + 95 = 0
Thus, our quadratic function that models the height of the object is y = -3.8(x - 5)^2 + 95.
The Problem:
Create a quadratic function to model the height of an object thrown into the air with the following conditions
The maximum height is between 90 and 100 m
The object is starts on the ground
The object takes at least 10 s before returning to the ground
1 mark for providing data
1 mark for completing an equation with a "maximum" between 90 to 100.
1 mark for showing work for the calculation of the maximum
1 mark for giving an equation that has the object starting on the ground
1 mark for showing work for the calculation of the initial height
1 mark for giving an equation that has a second zero of at least 10
1 mark for showing work for the second zero
1 answer
1 answer