Math Help - Parabola

1. Parabola

What is the maximum height that an 8-foot-wide truck can be in order to pass through a parabolic tunnel that is 18 feet high and 12 feet wide?

I drew the parabola, and tried to find similar triangles, but i didn't notice anything. Any help is appreciated.

2. Well here's a parabola that has x-intercepts 0, 12.

y = $a(x^2-12x)$ , you should find 'a' given the height is 18, i.e. use the point (6,18)

What do you get?

3. Hello, rtblue!

What is the maximum height that an 8-foot-wide truck can be in order
to pass through a parabolic tunnel that is 18 feet high and 12 feet wide?
Code:
        (0,18)|
*
*  |  *
*    |    *
*     |     *
|
*      |      *
|
|
----*-------+-------*----
(-6,0)     |     (6,0)
|

The general equation of a parabola is: . $y \:=\:ax^2 + bx + c$

The parabola passes through three points: . $(6,0),\:(\text{-}6,0),\:(0,18)$

Substitute these coordinates into the equation:

. . $\begin{array}{ccccccc}
(6,0)\!: & 36a + 6b + c &=& 0 \\
(\text{-}6,0)\!: & 36b - 6b + c &=& 0 \\
(0,18)\!: & 0a + 0b + c &=& 18 \end{array}$

Solve the system of equations: . $a = \text{-}\frac{1}{2},\;b = 0,\;c = 18$

Hence, the parabola is: . $y \;=\;\text{-}\frac{1}{2}x^2 + 18$

The truck is 8 feet wide.

How high is the tunnel at $x = 4$ ?

Code:
            18|
*
*  |  *
*- - + - -*
*|    |    |*
|    |    |
* |    |   y| *
|    |    |
|    |    |
----*--+----+----+--*----
-6 -4    |    4  6
|

At $x = 4\!:\;y \:=\:\text{-}\frac{1}{2}(4^2) + 18 \:=\:10$

Therefore, the truck can be 10 feet high (at most).

4. with axis of symetry on y axis

thot I would join the party altho the high cards have already been dealt

but...

you could view this as an parabola with axis of symetry on y axis. thus the vertex is $(0,18)$ and we have another point on the graph of $(6,0)$ not using the the $(-6,0)$ point

then

vertex $= (h,k)$ or $(0,18)$

$(x-h)^2 = -4a(y-k)\Rightarrow x^2 = -4a(y-18)$

from the other point $(6,0)$ we plug in to get $a$
$6^2 = 4a(0-18)$ so $a=-\frac{1}{2}$

so the equation for axis of symmetry over y is

$x^2 = -2(y-18)$

leting $x=4$then $y=10$ or $10ft$ but this of course actually touches the roof of the tunnel.

5. The problem was not about "similar triangles", it was about the largest rectangle (the truck) that could be fitted into the parabola.

6. vertex $\displaystyle \frac{-b}{2a}$ lets assume the parabola crosses at $\pm 12$ on the x axis and the y intercept is at (0,18). Now, first you need to find the equation of the parabola. Then set the parabola equal to the equation x=8.

7. Write the equation of your parabola in turning point form...

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

You know the turning point is $\displaystyle (6, 18)$ (since the parabola is symmetrical), and you know that $\displaystyle (0,0)$ and $\displaystyle (12,0)$ are points. Substitute into the equation...

$\displaystyle 0 = a(12 - 6)^2 + 18$

$\displaystyle 0 = 36a + 18$

$\displaystyle -18 = 36a$

$\displaystyle -\frac{1}{2} = a$.

So the equation of the parabola is $\displaystyle y = -\frac{1}{2}(x - 6)^2 + 18$.

Your truck is $\displaystyle 8\,\textrm{ft}$ wide, so that means there are $\displaystyle 4\,\textrm{ft}$ on either side. So you want to find the height of your parabola ( $\displaystyle y$ value) when $\displaystyle x = 2$ or when $\displaystyle x = 10$.