Question

The stream of water from a fountain follows a parabolic path. The stream reaches a maximum height of 7 feet, represented by a vertex of (4,7)
, and lands 8 feet from the water jet, represented by (8,0)
. Write a function in vertex form that models the path of the stream.

Answer 1

First of all, I'm going to assume that we have a concave down parabola, because the stream of water is subjected to gravity.

If we need the vertex to be at $x=4$, the equation will contain a $(x-4)^2$ term.

If we start with $y=-(x-4)^2$ we have a parabola, concave down, with vertex at $x=4$ and a maximum of 0.

So, if we add 7, we will translate the function vertically up 7 units, so that the new maximum will be $(4, 7)$

We have

$y = -(x-4)+7$

Now we only have to fix the fact that this parabola doesn't land at $(8,0)$, because our parabola is too "narrow". We can work on that by multiplying the squared parenthesis by a certain coefficient: we want

$y = a(x-4)^2+7$

such that:

• $a<0$
• when we plug $x=8$, we have $y=0$

Plugging these values gets us

$0 = a(8-4)^2+7 \iff 16a+7=0 \iff a = -\dfrac{7}{16}$

As you can see in the attached figure, the parabola we get satisfies all the requests.

Answer 2

The path followed by the stream in vertex form is $y = -\frac{7}{16}(x - 4)^2 + 7$

The equation of a parabola is represented as:

$y = a(x - h)^2 + k$

The given parameters are:

$(h,k) =(4,7)$ -- the vertex

$(x,y) =(8,0)$ -- the point it passes through

Substitute $(h,k) =(4,7)$ in $y = a(x - h)^2 + k$

$y = a(x - 4)^2 + 7$

Substitute $(x,y) =(8,0)$ in $y = a(x - 4)^2 + 7$

$0 = a(8 - 4)^2 + 7$

$0 = a(4)^2 + 7$

$0 = 16a+ 7$

Collect like terms

$16a= - 7$

Make a, the subject

$a = -\frac{7}{16}$

Substitute $a = -\frac{7}{16}$ in $y = a(x - 4)^2 + 7$

$y = -\frac{7}{16}(x - 4)^2 + 7$

Hence, the equation of the parabola is: $y = -\frac{7}{16}(x - 4)^2 + 7$

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