a Substitute the coordinates of the vertex into the given equation. Then, use the fact that the flight path passes through the origin to make the equation.
B
b Make a plot to find the domain and range of the function. In the graph, the x-axis represents the water level.
C
c Use the same process you did in part A. Notice that the flight path still passes through the origin.
A
a y=-5/1089(x - 33)^2 + 5
B
b The domain represents the horizontal distance covered by the fish out of the water, while the range is the height of the fish above the water.
C
c Yes, it changes to - 1225. Although the vertex changes, the graph still passes through the origin and it causes a horizontal shrink or stretch, which changes the value of a.
Practice makes perfect
a We need to write an equation using the form y= a(x- h)^2+ k, where ( h, k) are the coordinates of the vertex. Since we are told that (33,5) is the vertex of the flight path, we can substitute it to get an initial equation.
y= a(x- 33)^2+ 5To find a, we will use the fact that the path passes through (0,0). Let's substitute (0,0) into the equation above and solve for a.
Next, substituting all the constants into the initial equation, we get the flight path of the fish.
y= -5/1089(x- 33)^2+ 5
b Let's plot the function to see, graphically, the domain and range of the function.
From the graph we see that the domain is 0 ≤ x ≤ 66 and the range is 0 ≤ y ≤ 5.
The domain represents the horizontal distance covered by the fish out of the water, while the range is the height of the fish above water.
c Yes. The value of a changes. Indeed, if the new vertex is (30,4) then the initial equation is y= a(x-30)+4. As in part A, to find a we will substitute (0,0) into this equation.
