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#1 2021-05-01 10:30:25

mathland
Member
Registered: 2021-03-25
Posts: 444

Arch Bridge

The arch support of a bridge is modeled by y = −0.0012x^2 + 300, where x and y are measured in feet and the x-axis represents the ground.

A. Find one x-intercept of the graph.

B. Explain how to use the intercept and the symmetry of the graph to
find the width of the arch support.

For part A, I simplyet y = 0 and solve for x. Yes?

For part B, I need to graph the equation first.
How do I use the intercept and the symmetry of the graph to
find the width of the arch support.

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#2 2021-05-01 11:49:07

zetafunc
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Registered: 2014-05-21
Posts: 2,432
Website

Re: Arch Bridge

Yes is the answer to your first question.

For your second, have a think about the shape of your graph. What does it look like? And what do you think might represent the 'arch support' here?

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#3 2021-05-01 20:23:19

Bob
Administrator
Registered: 2010-06-20
Posts: 10,167

Re: Arch Bridge

This quadratic has y=0 as a line of symmetry.  When x = 0 y = 300.  f(x) = f(-x) for all x and, as the squared term is negative, the quadratic curve is 'n' shaped rather than 'u' shaped.

So the points where the curve cuts the x axis (two) are the points on the ground where the arch starts (I'm ignoring any brickwork below ground level).

So if you can find these points you can easily say how far apart they are.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2021-05-02 06:41:43

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Arch Bridge

zetafunc wrote:

Yes is the answer to your first question.

For your second, have a think about the shape of your graph. What does it look like? And what do you think might represent the 'arch support' here?

The shape of an arch is typically a parabola.  All quadratic form a parabola shape. How do I use the intercept and the symmetry of the graph to
find the width of the arch support.

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#5 2021-05-02 19:41:09

Bob
Administrator
Registered: 2010-06-20
Posts: 10,167

Re: Arch Bridge

post #3 ?


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2021-05-03 09:37:09

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Arch Bridge

Bob wrote:

post #3 ?

What do you mean by post # 3?

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#7 2021-05-03 10:39:49

Bob
Administrator
Registered: 2010-06-20
Posts: 10,167

Re: Arch Bridge

I'm viewing the forum using a Lenovo ideapad laptop.  The screen is roughly 14 inches wide.  In the top left hand corner of every post is a number: #1 for the first post; #2 for the next and so on.  If you cannot see this, what device are you using?  A mobile screen is probably too small for you to make much sense of many of my replies.

In the third post of this thread I explained exactly how to work out the width.  Try graphing the equation.

It's too late at night for me to look now at any of your other posts.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2021-05-03 11:22:08

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Arch Bridge

Bob wrote:

I'm viewing the forum using a Lenovo ideapad laptop.  The screen is roughly 14 inches wide.  In the top left hand corner of every post is a number: #1 for the first post; #2 for the next and so on.  If you cannot see this, what device are you using?  A mobile screen is probably too small for you to make much sense of many of my replies.

In the third post of this thread I explained exactly how to work out the width.  Try graphing the equation.

It's too late at night for me to look now at any of your other posts.

Bob

1. I use my cell phone for most of my threads. I don't have a computer or laptop.

2. I will graph the equation and return here if needed to continue our discussion.

3. You are not obligated to look at my posts on a daily basis, Bob. Don't feel pressured to reply daily. Reply when time allows. Math is a just a hobby for me at 56 years old. Ok?

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