The path of the fireworks will be in the form of a parabola, whose maxima is 30. We know that the vertex of a downward parabola gives maxima. So our parabola will be a downward opening parabola.

Are you familiar with the vertex form of a parabola?

** The vertex form...**

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The path of the fireworks will be in the form of a parabola, whose maxima is 30. We know that the vertex of a downward parabola gives maxima. So our parabola will be a downward opening parabola.

Are you familiar with the vertex form of a parabola?

**The vertex form of a parabola whose vertex is at ( h, k) is given by `y=a(x-h)^2 +k` . For a downward parabola, the value of a must be negative.**

Let us assume that the boat is at origin (0,0).

The zeros of the parabola will be at (0,0) and (100,0) because the span of the parabola should be 100 meters.

We know that a parabola is symmetric along the vertical line that passes through the vertex. We can use this information to find the *x*-coordinate of the vertex, which will be the mid-point of 0 and 100.

We know that 50 is in the middle of 0 and 100, so the *x*-coordinate of the vertex is 50. We already know that the *y*-coordinate of the vertex is 30, because that is the height that the fireworks must reach. Therefore, the vertex is at point (50,30). I have attached an image of the path of the fireworks.

Using those values of *h* and *k*, we get:

`y=a(x-50)^2+30`

Our next step is to find the value of *a*. We have two points on the *x*-axis, and we can use either point to solve for *a*. Please go ahead and use the coordinates of the point (0,0) for *x* and *y* and solve for *a* before looking at the next step.

`0=a(0-50)^2+30`

`0=2500a+30`

`0-30=2500a+30-30`

`-30=2500a`

`(-30)/2500=a`

`(-3)/(250)=a`

Upon substituting the value of *a* in our equation, we get:

`y=(-3)/(250)(x-50)^2+30`

Therefore, the above equation will model the path of the fireworks.