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# Water is spraying from a nozzle in a fountain, forming a parabolic path as it travels through the air. The nozzle is 10 cm above the surface

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Water is spraying from a nozzle in a fountain, forming a parabolic path as it travels through the air. The nozzle is 10 cm above the surface of the water. The water achieves a maximum height of 100 cm above the water’s surface and lands in the pool. The water spray is again 10 cm above the surface of the water when it is 120 cm horizontally from the nozzle. Write the quadratic function in vertex form to represent the path of the water if the origin is at the surface of the water directly below the nozzle.

Guest Oct 20, 2014

#1
+91404
+10

You need to draw the picture first

you have a parabola that goes through the points (0,10), (120,10) with the vertex at (60,90)

since you have the vertex, you only need to use one of the other points.

(h,k) is the vertex so h=60, k=90

\$\$\\(x-h)^2=4a(y-k)\\\\
(x-60)^2=4a(y-100)\\\\
\$Sub in x=0, y=10 \$\\\\
(0-60)^2=4a(10-100)\\\\
3600=4a(-90)\\\\
900=a(-90)\\\\
a=-10\\\\
\$So the equation is \$\\\\
(x-60)^2=-40(y-100)\\\\\$\$

I suppose that is vertex form.

check

https://www.desmos.com/calculator/qim1rgkyfm

Melody  Oct 20, 2014
Sort:

#1
+91404
+10

You need to draw the picture first

you have a parabola that goes through the points (0,10), (120,10) with the vertex at (60,90)

since you have the vertex, you only need to use one of the other points.

(h,k) is the vertex so h=60, k=90

\$\$\\(x-h)^2=4a(y-k)\\\\
(x-60)^2=4a(y-100)\\\\
\$Sub in x=0, y=10 \$\\\\
(0-60)^2=4a(10-100)\\\\
3600=4a(-90)\\\\
900=a(-90)\\\\
a=-10\\\\
\$So the equation is \$\\\\
(x-60)^2=-40(y-100)\\\\\$\$

I suppose that is vertex form.

check

https://www.desmos.com/calculator/qim1rgkyfm

Melody  Oct 20, 2014
#2
+80776
+5

We have that

y = ax^2 + bx + c     we know that the point (0, 10) is on the graph

And we have another point on the graph  (120, 10)...so we have

10 = a(120)^2 + b(120) + 10 →  0 = 14400a + 120b

We also know that the parbola is symmetric....so the max occurs at x =  (120 + 0) / 2 = 60

And the max height = 100....so the point (60, 100) is also on the graph...so we have

100 = a(60)^2 +b(60) + 10 → 90 = 3600a + 60b

So using....

14400a + 120b = 0  and

3600a + 60b = 90

Multiply the second equation by -2 and add it to the first....this gives

7200a = -180

a = -1/40

So, solving for b we have  ...... 3600(-1/40) + 60b = 90 →  -90 +60b = 90 → 60b = 180 →  b = 3

So....our function is

y =(-1/40)x^2 + 3x + 10

You can see the graph here.....https://www.desmos.com/calculator/zqusuapft0

I forgot that you wanted this in vertex form......we have......

-40y = x^2 - 120x - 400 →

-40y + 400 = x^2 - 120x  .... complete the square.....

-40y + 400  + 3600 = x^2 - 120x + 3600 → (factor both sides)

-40(y - 100) = (x - 60)^2

CPhill  Oct 20, 2014

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