Can a Trailer Fit Under a Parabolic Arch Bridge?

[mustang]

Problem 11.
A hauing company needs to determine whether a large house trailer can be moved along a highway that passes under a bridge with an opening in the shape of a parabolic arch, 12m wide at the base and 6 m high in the center. If the trailer is 9m wide and 3.2 m tall(measured from the ground to the top of the trailer), will it fit under the bridge?

 

[Integral]

You need to come up with a formula for the parabola.
Let the origin be the point on the ground below the center of the arch.

You should know that the formula for the parabola will have the form
[tex] y(x)= ax^2 +b [/tex]
We need to determine the values of a and b.

We know that y(0)=6 and y(6)=0 (This is from your description of the parabola)

Using the first we find
[tex] y(0)= a(0)^2 + b = 6 [/tex]
or
b=6

Using the second we find
[tex] y(6)=a (6)^2 + 6 =0 [/tex]
[tex] a = - \frac 1 6 [/tex]

so we have for the parabola

[tex] y(x)= - \frac {x^2} 6 + 6 [/tex]

Now you just need to evaluate this function at the point corresponding to the edge of the truck (assume that the center of the truck is at 0)

Can you finish it?

 

[ShawnD]

First make a formula for the parabola.

It's 6 high at the centre so b is 6. The roots of the parabola (when it touches the ground) will be at -6,0 and 6,0.
Combine those roots together:

[tex](x + 6)(x - 6) = x^2 - 36[/tex]

Since the parabola goes down, multiply that function by -1

[tex]-1(x^2 - 36) = -x^2 + 36[/tex]

The middle of the parabola is supposed to be 6 high, not 36. Divide the function by 6.

[tex]\frac{-x^2 + 36}{6} = \frac{-1}{6}x^2 + 6[/tex]

Now check that function with your calculator. You can see that when x is -6, y is 0. When x is 0, y is 6. When x is 6, y is 0.

Now that you have the function for the arch, the question should be fairly striaght forward. Since the trailor is 9 wide, check what the height of the arch is at x = -4.5 and x = 4.5
Is the arch higher or lower than the trailor at those x positions?

 

[mustang]

It won't fit!

I used the formula a(x-r1)(x-r2) where I found a=-64/405. I placed it in the formula y=ax^2+b and got b=12.2. So the house won't fit.

I also some other problems that I need help on:
9. Find an equation of the quadratic function described:
d. The minimum value of f is f(3)=-5, and f(10 =2.
Problem 15.
(a) suppose that a is a positive integer such that ax^2+x-6 can be factored. Find the five smallest value of a. (b) Find two values of a that are greater than 100.