Applications of Polynomial Functions (Word Problem)

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plur222

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We have just started this chapter, and I was grouped with 2 others that didn't want to work 'together', so the questions were divided up ... this is mine ... I have attempted it, and done the initial graph - but don't know what formula to start with, and don't know what they are asking when they say "y-axis bisects the highway and the parabolic arch and the x-axis is placed at the base of the parabola".

Any help or pointers would be muchly appreciated!

A wildlife corridor must be designed to arch over a major highway. The arch over the highway is to be constructed in the shape of a parabola. The highway is 8m wide and will be centered under the parabolic arch. the arch must span a total width of 16m at it's base. the arch must provide a minimum clearance of 4m over the highway.

a) Sketch the parabolic arch and highway, labeling all relevant dimentions

b) Place a coordinate system on your sketch above so that the y-axis bisects the highway and the parabolic arch and the x-axis is placed at the base of the parabola. Find the quadratic function in standard form that describes the height of the arch, y, as a function of the horizontal distance from the center, x. Show all your steps.

c) State the maximum height of the arch over the highway (approximate to the nearest tenth of a meter)[/b]
 
plur222 said:
... don't know what they are asking when they say "y-axis bisects the highway and the parabolic arch and the x-axis is placed at the base of the parabola".
Click to expand...

This is telling you where on the xy-plane to place your diagram.

The x-axis is the ground.

The y-axis goes through the center of the highway.

The word "bisect" is a verb. It means to cut something into two equal pieces.

I will upload an image in a few minutes ...
 
Does your drawing look anything like the top illustration, below?

The bottom illustration is after the intoduction of an xy-coordinate system, as instructed.

(Double-click to expand the image.)

[attachment=0:4nvil788]wildlife_arch.JPG[/attachment:4nvil788]

You have the coordinates of both x-intercepts, as well as the coordinates of the vertex point. This is sufficient to find the equation whose graph is this parabola.

I note that the height above the center of the highway must be at least four meters. This means that it could be more, so your diagram might differ from mine. I chose to make the height exactly four meters. (Four meters is about 13.1 feet; this does not leave much clearance for tall loads.)

You can choose to make your maximum height more than 4, if you like.

Let us know if you need more help. Please show your work and explain what you're thinking, so that we can determine where to continue helping you.

 

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That is exactly what my graph looked like - so that's a good sign!

I tried using transformations to get a function out of it:

Y = x^2 (-2, 4) ( 0, 0) ( 2, 4)
Y1 = -1(x^2) (-2,-4) ( 0, 0) ( 2,-4)
Y2 = -1(x^2)+4 (-2, 0) ( 0, 0) (2, 0)
Y3 = -1(1/4x^2)+4 (-8, 0) ( 0, 4) ( 8, 0)

But is it -1(1/4x^2) + 4

Or -1(1/4x)^2 + 4

I'm not sure I should have had to use transformations, if not that - I'm not sure what formula to use ...?...

I think I have come up with Standard form f(x) = a(x-h)^2 + k

They also want me to find the maximum value for the height of the arch, do I test values for that? or could it be infinite?
 
A wildlife corridor must be designed to arch over a major highway. The arch over the highway is to be constructed in the shape of a parabola. The highway is 8m wide and will be centered under the parabolic arch. the arch must span a total width of 16m at it's base. the arch must provide a minimum clearance of 4m over the highway.

a) Sketch the parabolic arch and highway, labeling all relevant dimentions

b) Place a coordinate system on your sketch above so that the y-axis bisects the highway and the parabolic arch and the x-axis is placed at the base of the parabola. Find the quadratic function in standard form that describes the height of the arch, y, as a function of the horizontal distance from the center, x. Show all your steps.

c) State the maximum height of the arch over the highway (approximate to the nearest tenth of a meter)


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I have the graph, but don't know how to go about creating a formula

I've attempted using Transformations and gotten f(x) = -1(1/4 x)^2 + 4 ....... but don't know if it should be -1(1/4 x)^2 + 4 ........ ?

I think I have to come up with Standard form though which is f(x) = a(x-h)^2 + k

Do I use trial and error to find the maximum value for the height of the arch?

HELP!
 
plur222 said:
A wildlife corridor must be designed to arch over a major highway. The arch over the highway is to be constructed in the shape of a parabola. The highway is 8m wide and will be centered under the parabolic arch. the arch must span a total width of 16m at it's base. the arch must provide a minimum clearance of 4m over the highway.

b) Place a coordinate system on your sketch above so that the y-axis bisects the highway and the parabolic arch and the x-axis is placed at the base of the parabola. Find the quadratic function in standard form that describes the height of the arch, y, as a function of the horizontal distance from the center, x. Show all your steps.
Click to expand...
These requirements state that the vertex will be at (h, k) = (0, k), where you still need to find the maximum height "k". The x-intercepts will be at x = -8 and x = 8, and you need the height to be y > 4 at x = -4 and x = 4. For simplicity, let y = 4 and x = 4.

This gives you the equation y = ax[sup:2xkl3jeq]2[/sup:2xkl3jeq] + k and two points, (4, 4) and (8, 0). Plug these points into the equation and solve for "a" and "k". The value of "k" will give you the answer to part (c).

:D
 
plur222 said:
… (-8, 0) … ( 8, 0)

But is it -1(1/4x^2) + 4 Nope. This one does not give y=0 when x is -8 or 8.

Or -1(1/4x)^2 + 4 <<< It's this one because this one gives zero when x is -8 or 8.

I'm not sure I should have had to use transformations, … There are other methods (see below).

I think I have come up with Standard form f(x) = a(x-h)^2 + k This is easy to do (see below).

They also want me to find the maximum value for the height of the arch, do I test values for that? I'm not sure why they ask you to find the height. They previously implied that you need to pick an arbitrary height.

or could it be infinite? Infinite height? How could wildlife use such a path to cross the highway?!!
Click to expand...


Write your function like this:

f(x) = -(x/4)^2 + 4

OR

f(x) = -([1/4]x)^2 + 4

In other words, do not write a -1 in front of the parentheses. When a coefficient is 1, we do not write it. When a coefficient is -1, we only write the negative sign.

When typing fractions using a keyboard, we often need grouping symbols to clearly distinguish what's in the numerator and what's in the denominator.

Typing 1/4x^2 is ambiguous. There is no way to tell if this means 1/4 times x^2 VERSUS 1 divided by 4x^2.

That's why I put square brackets around the coefficient 1/4 above.

I see nothing wrong with using the logic of transformations to find the equation, but I would have approached it differently. (There are at least four methods.)

All quadratics may be written in this form:

y = Ax^2 + Bx + C

Now, think about the vertex point (0, 4).

Clearly, C is the value of y when x is zero. We know that y is 4 when x is zero. Therefore, without doing any calculations, we realize that C = 4.

y = Ax^2 + Bx + 4

We have two more pairs of (x, y) values that satisfy this equation: (-8, 0) and (8, 0).

If we substitute these x- and y-values into this equation, we end up with a system of two equations in A and B.

0 = 64A - 8B + 4

0 = 64A + 8B + 4

We can add these two equations to solve this system.

0 = 128A + 8

A = -1/16 and B = 0

This gives us our quadratic equation:

y = -(1/16)x^2 + 4

It's easy to put y = Ax^2 + Bx + C into standard form. We usually complete the square, but, with B = 0, we can do it mentally.

y = A(x - h)^2 + k

Do you know? (h, k) are the coordinates of the vertex point.

h = 0 and k = 4

y = A(x - 0)^2 + 4

y = Ax^2 + 4

We already know that A = -1/16; therefore, we see that we actually had arrived at standard form, above.

y = -(1/16)x^2 + 4

(If you're into transformations, then you could also realize that h must be zero because there is no horizontal shift. There is no horizontal shift because the instructions indirectly tell us that the y-axis corresponds to the parabola's axis of symmetry.)

As far as part (C) in this exercise is concerned, I'm not quite sure what to think.

The instructions told you to build an arch whose height is "at least" four meters. This tells me that you're free to pick your own height at the very beginning. I don't know why they want you to calculate a height (to the nearest tenth of a meter, no less) when you get to choose the exact height to begin with. I might be misreading this part of the exercise.

You could choose to make the height 4.1 meters. If so, then the answer to part (C) is 4.1 meters. That's not an estimate!

You could choose to make the height 4.4 meters. If so, then the answer to part (C) is 4.4 meters. That's not an estimate!

You could choose to make the height 5.7 meters. If so, then the answer to part (C) is 5.7 meters. That's not an estimate!

Clearly, we want wildlife to be able to navigate over the arch. The entire purpose of building this arch is for "chickens to cross the road". Since the two locations where the arch meets the ground are fixed, increasing the height causes the slope to increase at both ends of the arch. Increase the height too much, and the slope will be too steep to climb for the types of animals that would damage our vehicles when struck at highway velocities.

With the maximum height at exactly four meters, the slope at the base is 1. In other words, the base enters the ground at a 45-degree angle. That seems a wee bit too steep to entice animals to use an arch versus walking on the highway, but I guess it's reasonable for an algebra application that's cooked-up to use quadratic modeling.

If the height were to be six meters, then the base would enter the ground at more than 56 degrees. That slope is close to unreasonable.

Maybe, somebody else at this web site will chime in, regarding part (C). I just don't understand why an instructor would ask you to estimate an arbitrary value that they previous implied for you to choose on your own.

 
How do you get that you want to make y = 4 and x = 4 ?
Does k = 32 ?
 
Use your given data and \(\displaystyle y=a(x-h)^{2}+k\). No, no need for trial and error.

A constraint is that it has to have a minimum of 4 feet clearance over the road. That will be at the edge of the road on both sides.

At coordinates (4,4) and (-4,4). Those are 'passing through' points.

The vertex is at (0,k). k is how high the parabola is at the center.

Where the parabola crosses the x-axis is also given as (8,0) and (-8,0).

That is plenty of info to use in the above formula and solve for a and k.

One would be \(\displaystyle 4=a(4-0)^{2}+k\)

Build another by using another of the points and solve the little, easily solved system you obtain.

Here is a graph of your parabola. I omit the equation. You can get that now.
 

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galactus said:
… A constraint is that it has to have a minimum of 4 feet clearance over the road. That will be at the edge of the road on both sides…
Click to expand...


(Hmmm, I'm getting that deja vue sensation, again. I believe that I contributed to this same discussion somewhere else.)

I interpreted the clearance over the road at its maximum (about 13.1 feet at the center).

Galactus interprets it differently (about 13.1 feet at the lateral edges of the road).

In hindsight, Galactus' interpretation makes more sense because doing so means that the maximum height is not known initially. In this case, asking the question in part (C) makes sense.

(I will look for that duplicate discussion, and post a link to this thread.)

 
This discussion is a duplicate thread.

Galactus responded to this discussion at the other location, with an interpretation of the phrase "the arch must provide a minimum clearance of 4m over the highway" that is better than what I described in my post above.

Galactus reads it as a 4-meter clearance at the lowest point directly above the outside edges of the road; I was reading it to mean a 4-meter clearance at the highest point (above the center of the road).

I believe that Galactus is right, and I was wrong. Therefore, most of the information that I posted above is wrong, too.

Please go to the other location HERE, and read Galactus' post.

Oh, it looks like Elizabeth got it right, too. Credit goes to her, also.
 
OMG - I don't understand why I can't get this - I am wasting hours and getting no where

How do you figure out what anything is with 4 = a(4-0)^2 + k ?

How do I find a ?

I understand the graph, but not the formulas and how to determine for a(x-h)^2 + k what the values of a, x, and k are, I know h=0 because that is the axis of symmetry and the x intercept value for the vertex point (0,k)

I have the x-intercepts as (-8,0) (8,0)
I have the points I can use as (-4,4) (4,4)

You guys say to "just plug in points" but what points do I plug in where?!?!

I don't even know what parts I'm not getting and what to look up on other help sites.......
 
plur222 said:
How do you figure out what anything is with 4 = a(4-0)^2 + k ? You don't. This is only one of two equations that you need to form.

Did anybody teach you about solving systems of equations? That usually comes before assigning an exercise such as this one.

(I posted the steps in forming of a system of equations when you still had this discussion in two different places. I showed you one method for solving that system. I do not know whether or not you read it or understand it.)

You guys say to "just plug in points" but what points do I plug in where?!?!

You substitute the x-coordinate for x, and you subsitute the y-coordinate for y.
Click to expand...

Galactus gave you a formula for y:

y = a(x)^2 + k

Do you see the symbol x? Replace it with the number 4.

Do you see the symbol y? Replace it with the number 4.

We can do this because we know that the point (4, 4) satisfies the parabola's equation, and we know that (4, 4) means x = 4 and y = 4. These substitutions give us ONE equation containing the symbols a and k.

4 = a(4)^2 + k

It simplifies.

16a + k = 4

We don't stop here.

We need two equations with the symbols a and k, in order to find their values.

If we were to now use the point (-4, 4) to write the second equation, we would end up with the same equation as we did with (4, 4). That's no good; we need two different equations.

So, use the point (8, 0) instead. Substitute the number 8 for x and the number 0 for y.

0 = a(8)^2 + k

It simplifies.

64a + k = 0

Now we have a "system of two equations" containing the two symbols a and k.

16a + k = 4
64a + k = 0

What do you know about solving systems of equations?

 
This is what I have come up with
I'm feeling slightly less crazy
Am I on the right track?
I thought "a" would = -1/4 .... not -1/12

I really appreciate everyone's help - this site is a godsent!

I will be going back and studying further - but it would kill me to know I have let the two other people in my group down tonight.
 
Your worksheet indicates you got a = -1/12, which is correct; so what are you asking?

By the way, you can solve such equations much quicker this way (and save on paper!):
16a + k = 4 [1]
64a + k = 0 [2]

Subtract the equations; [1] - [2]:
-48a = 4
a = 4 / -48
a = -1/12
 
My final equation has to be in the format: a(x-h)^2 + k

I was told the final equation would be y = - (x/4)^2 + 4 which would mean a = 1/4 yes?

Is there a way I can check my work, assuming the following is correct:

a = - 1/12

k = - 16/3

y = 0

Which x value would I put into the equation? Or I suppose -8 and 8 would give me the same since it's being squared.
 
plur222 said:
I was told the final equation would be y = - (x/4)^2 + 4 which would mean a = 1/4 yes?
Click to expand...


That equation models an arch whose vertex is 4 meters above the center of the highway.

We changed the interpretation (i.e., the 4-meter height is no longer at the vertex), so the values of a and k changed.

Not that it matters for this exercise, but -(x/4)^2 would not mean that a = 1/4 because the 4 is being squared and there is a negative sign in front. It would mean that a = -1/16.

-
Is there a way I can check my work, assuming the following is correct:

a = - 1/12

k = - 16/3
Click to expand...


You got the right values on your latest work shown above, but it looks like you made a typographical error here because k is not negative.

a = -1/12

k = 16/3

The equation that models this arch is:

y = (-1/12)x^2 + 16/3

You can check this equation by making sure that you get 0 for y when you let x be 8 or -8. Make sure that you get 4 for y when you let x be 4 or -4. You could check to make sure that you get 16/3 for y when you let x be zero, but that's obvious, right?

-
I suppose -8 and 8 would give me the same since it's being squared.
Click to expand...


Yup, yup.

Letting x be -4 and 4 leads to the same situation. This is all because of symmetry.

 
GOOD! DONE! Huge thanks to everyone that helped - I have studying to do, I'm pretty sure they won't give me 11 hours on the exam ... lol
:D
 
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