# Thread: Water slide~ Parabola question

1. ## Water slide~ Parabola question

It's kind of a word problem that includes a graph, however it prominently just loses me completely.

Andrew has designed a new water slide as shown below. H and x in metres. It is constructed using parts of two parabolas that intersect at the point when x=8
the slide starts flat and finishes flat at the point (20,0.5) and passes through two other points(4,34.5) and (14, 5)

Determine the height of the top of the slide, by modeling the slide with two parabolas. THat is, find the h-intercept of the graph.

Help appreciated.

2. Parabolas are curves, not "flat" lines. So I think we're going to need to see the picture before we can advise...?

3. Its constructed of parts of two parabolas. So start with y = -x^2 on the top and halfway switch to y=x^2 so you end at lowest part of that graph. Now you start flat and end also flat.

4. uhmmm i think its the maxium and minimum points...

5. Originally Posted by Mooki
It's kind of a word problem that includes a graph, however it prominently just loses me completely.

Andrew has designed a new water slide as shown below. H and x in metres. It is constructed using parts of two parabolas that intersect at the point when x=8
the slide starts flat and finishes flat at the point (20,0.5) and passes through two other points(4,34.5) and (14, 5)

Determine the height of the top of the slide, by modeling the slide with two parabolas. THat is, find the h-intercept of the graph.

Help appreciated.
There was no image attached, so I made one. [See image below]

It has been several days since it was originally posted,
so the actual answer does not appear significant at this time.
The coordinates for P5 should be (-2, 43.5 )

However, in getting a solution a difficulty arose:

Parabolic Equation: [tex] y = ax^2 + bx + c [tex]
For the opening upwards parabola:
Points P1 and P2 were translated to the origin.
The value of "a" was determined to be 1/8.
The y-coordinate for P3 was obtained.
After re-translating back to P1(20,0.5) the values for "b" and "c" were obtained.

thus, for the parabola opening upwards:
$y = 0.125x^2 - 5x + 50.5$

The slope of the line tangent to point P3 is -3

For the parabola opening downward:
at P4: $34.5 = a 4^2 + 4b + c$
at P3: $18.5 = a 8^2 + 8b + c$
the difference
16 = -48a - 4b

The slope at P3 must be -3, so: $-3 = 16a +b$

thus 2 equations:
16 = -48a -4b
-3 = 16a + b

---
multiply 2nd Eqn by 3 and add:
16 = -48a -4b
-9 = 48a +3b
--------------
7 = - b
-7 = b
plugging back in, results in: a = +0.25
but a must be negative?
Why isn't the value for "a" negative?

---
multiply 2nd Eqn by 4 and add:
16 = -48a -4b
-12 = 64a +4b
--------------
4 = 16a
0.25 = a
Since a must be negative?
plugging (-0.25) back in, results in: b = -1

this works for the parabola opening downwards:
$y = -0.25x^2 - 1x + 42.5$

But I do not understand why "a" does not evaluate to a negative in the equations above.

.