Quadratics

5.2 Some Typical Textbook Questions

Here is a typical question on quadratics that hope-to-be mathematics teachers must answer on a state licensure exam. How would you fare with this?

Recall that all solutions appear in the COMPANION GUIDE to this QUADRATICS course.

 

PRACTICE 110: TEACHER LICENSURE TYPE-QUESTION

Geologists suspect that the cross-section shape of a newly discovered impact crater in the Australian outback can be well approximated by a quadratic curve. (Really? Hmm!) The crater is 50 feet wide.

They erect a platform across the crater and drop rope at ten foot intervals across the platform to measure the depth of the crater at these locations. The first rope, ten feet in from the rim of the crater, is 40 feet long.

Q5_2_pic1

Placing this diagram on a coordinate system of your choice, find a quadratic equation that fits these initial data values. .

Assuming that the crater is indeed quadratic, what is the depth of the crater?

What, according to your equation, are the lengths of the remaining three ropes?

To test their conjecture that the crater is indeed quadratic, two more ropes of lengths 50.4 feet are to be hung from the platform. Predict exactly where along the platform they should be placed so as to just touch the floor of the crater.

 

Here is a silly example that is pretending that insurance companies use quadratic equations.

 

EXAMPLE 111:

PRANG-AND-BANG Car Insurance has the following data for the average number of accidents folk of different ages have for every 100,000 km driven:

Q5_2_pic2

They are thinking of increasing their insurance coverage to drivers age 80 and need to predict the number of accidents per 100,000 km driven for this age-group. They suspect an increase in number so they feel that modeling this data with a quadratic formula would be appropriate.

a) Write down a quadratic formula for this data and give the average number of accidents it predicts per 100,000 km of driving for 80-year old drivers.

b) According to this model are 16-year-old drivers better or worse than 80-year-old drivers?

c) According to the model, which age group has the best driving record?

d) Suppose company only wishes to cover folk in the age ranges with 10 or less accidents per 100,000 km of driving. What ages will it then cover?

 

Answer: Let \(N(x)\) be the number of accidents per year for age \(x\).

a)

\(N(x)= 8.5\frac{(x-30)(x-50)}{(-10)(-30)}+3.7\frac{(x-20)(x-50)}{(10)(-20)}+1.3\frac{(x-20)(x-30)}{(30)(20)}\)

        \(=\frac{85}{3000}(x-30)(x-50)-\frac{37}{2000}(x-20)(x-50)+\frac{13}{6000}(x-20)(x-30)\).

 

It seems tempting to put everything over a common denominator of \(6000\).

\(N(x)=\frac{170(x-30)(x-50)-111(x-20)(x-50+13(x-20)(x-30)}{6000}\).

 

We have:

\(N(80)=\frac{170(50)(30)-111(60)(30)+13(60)(50)}{6000}\)

        \(=\frac{170\times15-111\times18+13\times30}{60} = \frac{942}{60}=15.7\)

 

b)

\(N(16)=\frac{170(-14)(-34)-111(-4)(-34)+13(-4)(-14)}{6000}=\frac{66552}{6000} \approx 11.1\).

16-year old drivers are safer drivers according to this model.

 

c) Okay .. We’re going to have to simplify our formula. I’m game!

 

\(N(x)=\frac{170(x^{2}-80x+1500)-111(x^{2}-70x+1000)+13(x^{2}-50x+600)}{6000}\).

 

The \(x^{2}\) terms are: \(\frac{170-111+13}{6000}x^{2}=\frac{72}{6000}x^{2}\).

The \(x\) terms are: \(\frac{170(-80)-111(-70)+13(50)}{6000}x=-\frac{6480}{6000}x\).

The constant terms are: \(\frac{170(1500)-111(1000)+13(600)}{60000)}=\frac{151800}{6000}\).

So

\(N(x)=\frac{72x^{2}-6480x+151800}{6000}\).

 

To see the interesting \(x\)-values rewrite this as:

\(N(x)=\frac{72x(x-90)+151800}{6000}\).

We see that \(x=0\) and \(x=90\) are interesting, which means the safest drivers (the lowest value) occurs for \(x=45\).

 

d) We need to look at

\(\frac{72x^{2}-6480x+151800}{6000}=10\).

 

Multiplying through by \(6000\) gives:

\(72x^{2}-6480x+151800=60000\).

Halving this yields something good and ready for the box method:

\(36x^{2}-3240x+75900=3000\).

Q5_2_pic3

Subtract \(3000\) from both sides:

\(36x^{2}-3240x+72900=27000\)

\((6x-270)^{2}=27000\)

\(6x-27- \approx 164.3\) or \( 6x-270 \approx -164.3\).

\(6x\approx 434.3\) or \(6x\approx 105.7\).

\(x\approx 72.4\)  or \(x\approx 17.6\).

They will insure between the ages 17.6 and 72.4. □

Phew!

Doing this work by hand is a bit silly. In this day and age it would be appropriate to type our original formula in part a) into a computer algebra system and let it do the calculations and the simplifications. But at least this example shows that it is possible to answer these questions by hand if forced to. It is not the concepts that are hard, it is the annoying arithmetic that is hard

 

Here’s a problem for you with less intensive arithmetic:

 

PRACTICE 112: A ball is kicked into the air from a position of 5 feet off the ground. Let \(H(t)\) represent the height of the ball at time \(t\) seconds, with \(t=0\) corresponding to the time of the kick.

At \(t=2\) the ball is 69 feet in the air and at time \(t=3\) it is 53 feet in the air.

Galileo suggested that the height of an object in the air subject to gravity follows a quadratic formula with respect to time. Let’s assume this is the case here. (Air resistance, wind, and the like, actually, ruin this ideal!)

a) Write a quadratic formula of the form \(H(t)=at^{2}+bt+c\) for the data given.

b) What is the maximum height of the ball?

c) When does the ball hit the ground?

 

EXAMPLE 113: Here is a picture of curve that looks like it could be quadratic.

Q5_2_pic4

To test this, I used a ruler to make the following measurements on the picture:

Q5_2_pic5

These measurements give the data:

Q5_2_5A

Using three of the data points – and I choose \(\left({0},{0}\right)\) and \(\left({1},{2.8}\right)\)and \(\left({4.2},{0}\right)\) as zeros are always easy to work with – we can write down a quadratic equation that fits them:

 

\(y=0\frac{(x-1)(x-4.2)}{(-1)(-4.2)}+2.8\frac{(x)(x-4.2)}{(1)(-3.2)}+0\frac{(x)(x-1)}{(4.2)(3.2)}\)

        \(=0-\frac{2.8}{3.2}x(x-4.2)+0\)

        \(=-\frac{7}{8}x(x-4.2)\)

 

If curve drawn really is quadratic, then putting in \(x=2\) should give a value close to \(y=3.4\). (I’ll allow some error for my measuring inaccuracy.)

\(y(2)=-\frac{7}{8}(2)(-2.2)=3.85\).

This seems a bit high for my liking. I don’t think this curve is actually quadratic.

 

COMMENT: One could also use difference methods described in part I of this course to analyse the shapes of U-shaped curves.

 

PRACTICE 114:
a) Many algebra books show a picture of the St. Louis Gateway Arch at the header of the quadratics chapter. Is the arch actually in the shape of a quadratic curve? Find a picture of the arch and make some measurements on it. Use the technique of example 113 of the difference methods of part I to test whether its curve follows a quadratic formula.

b) Galileo thought that all ropes or chains that hang between two poles make the shapes of quadratic curves. (The shape of power lines, the shape of ropes that surround sculptures in art museums, and so on.)

Q5_2_pic6

Is the shape of a hanging chain quadratic? Take a photograph of a hanging string or rope and make some measurements on it. What do you think?

 

Self Check: Question 19

Another self check!


Self Check: Question 20

Lucky us! A second question!

Self Check: Question 21

The final of all final self check questions!

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