chapter 4 — linear and quadratic functionsnofrillspublishing.com/math1141/book/5.pdf · 2009. 12....

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Chapter 4 — Linear and Quadratic Functions

We will concentrate on the quadratic functions in this chapter. These are second degree polynomial functions, the next step up in complexity from linear functions, or lines. This will give a glimpse of analytical techniques that we’ll continue to expand on throughout this course and into calculus. The last topic concerns a simple economic model related to the issue of continuity of a function that may resemble a quadratic function.

Section 4.3 — Quadratic Functions and Their Properties

Quadratic functions, which graph as parabolas, permit full analysis without calculus techniques, so they are a natural launching point for the analysis of all polynomial functions.

Topic 1: Maximum or minimum of a quadratic function.Topic 2: Graphing quadratic functions.

Practice Problems: 11 – 65 odd

Topic 1: Maximum or minimum of a quadratic function.

Quadratic functions can be organized into this form: , where a ≠ 0. As we’ve seen before, the graph of a quadratic function is the shape known as a parabola. Although a relatively simple shape, parabolas actually have more geometric features than meet the eye. If you take the College Trigo-nometry course then you will also learn about the focus, directrix, maybe even the latus rectum of a parabola. In this course it is enough to learn about the vertex and its role as the minimum or the maximum of a quadratic function.

On the next page I have drawn a generic quadratic function and labeled the items of current interest for us. Of course, there are other kinds of these parabolas. Some point downwards, some have no x–intercepts, and, of course, they may vary in shape from wide to thin. But all have a vertex, and all have a y–intercept. (Why?) The axis of symmetry of a quadratic function is always a vertical line passing through the vertex. Naturally, this is the line of symmetry for the parabola.

In the last section we looked at some examples of quadratic functions through the lens of function trans-formations. That knowledge will inform us about quadratic functions, but the main point of this section is to find the vertex of a parabola and to apply that to mathematical models. Your textbook uses a completing the square technique to transform a quadratic function into the form , from which we get the vertex as the point (h, k), similar to a process you learned about related to circles. This is a perfectly good technique, albeit trickier than that used with circles. I will derive an alternative method for finding the vertex that is simpler to use in practice.


196 Chapter 4 — Linear and Quadratic Functions

If I gave you the x–intercepts of a quadratic function would you be able to find the x–coordinate of the vertex? Once you understand the symmetry of a parabola it should be obvious, right? The x–coordinate of the vertex lies right in the middle of the x–intercepts, or roots, of the parabola. (I will use the term “roots” inter-changeably with “x-intercepts.”) Next question: How do you find the middle of these x–intercepts?

First of all, the roots of a quadratic function are the solutions to . (They are x–inter-cepts because the x–axis is where the function, or y, equals zero.) We have done that before. The roots may be expressed in the general way from the quadratic formula. These roots are the following.

and

The middle of two numbers is their average, right? We find the average, or middle, of two numbers by adding them together and dividing by 2. Here is what happens when we do that with these roots.

It was simpler than it looked. The square root parts of the quadratic formula canceled one another in the calcu-lations. This means that the average, or the x–coordinate of the vertex, is simply the quadratic formula without the square root part.

It is easy to show that this little formula will still locate the x–coordinate of the vertex even when the pa-rabola does not have x–intercepts. For example, if the parabola was entirely above the x–axis then there would be no real roots. In fact, the roots would be complex numbers, which cannot be depicted in the x–y plane. If we changed the constant c in such a quadratic function, that would have the effect of vertically shifting the parabola. We could choose c to be anything we like such that it would raise or lower the parabola so that it did have two real roots, in which case our little formula would still provide the x–coordinate of the vertex. There

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is no number c in this formula, so the determination of the x–coordinate of the vertex is independent of what c equals.

So now we have a simple technique for finding the vertex of a parabola. Once you have the x–coordinate, finding the y–coordinate is simply a matter of plugging the x–coordinate into the function to see what the output y–value equals, as in the following example.________________________________________

Find the vertex of .

In this quadratic function a = 2 and b = – 12. It doesn’t matter what c equals when determining the x–coordinate of the vertex.

Now we plug this into the function in order to calculate the y–coordinate of the vertex.

Now let’s write out the answer and we’re done.

It turns out that this vertex is the minimum of the quadratic function it comes from. You should recall from the discussion about function transformations that the basic parabola opens upward, with the vertex at the origin, (0, 0). The parabola above is a transformation of , which was stretched by a factor of 2, shifted 3 units to the right and 11 units down. Thus it should be equivalent to . It is easy to confirm that if you multiply out and simplify then you will get .

For comparison purposes I will show you the textbook technique using completing the square process. If you are my student I don’t care which technique you use, but just about everyone prefers the first method as being quicker, which I will stick to after this example.________________________________________

Find the vertex of .

In order to apply completing the square as a technique we must first make the coefficient of x2 equal to 1. We can do this by factoring the current coefficient, 2, out of the x terms. Note that I’ve left the constant term 7, alone for the moment, as well as left some space inside the parentheses.

Inside the parentheses we will now complete the square on x2 – 6x. This process, which is similar to the one we used in one type of circle problem, will create a perfect square binomial. We do so by taking half of the x term, – 6, to get – 3. We then square – 3 to get 9, and then both add and subtract 9 inside the parentheses.



As you can see, we ended up adding and subtracting 9 on the same side of the equation instead of adding 9 to both sides of the equation. We also did this within the parentheses. The 9’s could cancel, so we have not nu-merically altered the equation of the function.

What was gained? We will now take the – 9 out of the parentheses, which is why this is a little tricky. Everything inside the parentheses is multiplied by 2, so the – 9 in the parentheses comes out as – 18. What is left inside the parentheses factors as a perfect square binomial. After combining – 18 and 7 we now have the quadratic function written as a transformation of y = x2.

I will now return to using the method for finding the vertex. Our remaining issue in this topic is how do you determine if the vertex is the minimum or the maximum of a quadratic function? This is an easy question to answer in light of function transformations.

The leading coefficient of a polynomial function is the number multiplying the highest degree term. In the case of quadratic functions this would equal to a, the number multiplying the x2 term. Whenever you transform the basic quadratic function y = x2 this coefficient tells us both the vertical stretch, and whether there was a reflection across the x–axis based upon whether a is negative or positive. If a is positive, then the parabola points upwards and the vertex is a minimum. If a is negative, then the basic parabola is inverted through reflection across the x–axis and the vertex is now the maximum.

Thus in our example using we determined that the vertex is (3, – 11). Since the leading coefficient, 2, is positive, meaning that we have an upward pointing parabola, then we can also say that f (3) = – 11 is the minimum of the function.________________________________________Find the vertex of and determine what the maximum or minimum of the function is, whichever is appropriate.

Since the leading coefficient, – 1, is negative, the graph of the parabola opens downwards. Thus the vertex gives us the maximum. Technically speaking, the maximum of the function is not the vertex, but the y–coordinate of the vertex, the largest number the function can attain.

Let’s first find the x–coordinate of the vertex.

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Careful when plugging this negative number into this function. Too many students get a positive number in the first term under similar circumstances, but – x2 is always negative when x is a real number, regardless of sign. That is because x is squared first according to the order of operations, and then the negative sign is applied.

Now we can write the vertex. Be sure to include it in parentheses so that it is written as a coordinate point.

Again, this parabola is pointed downwards and has a maximum peak due to the reflection indicated by the leading coefficient.

Reinforcement Problems

1. Find the vertex of and determine what the maximum or minimum of the func-tion is, whichever is appropriate.

2. Find the vertex of and determine what the maximum or minimum of the function is, whichever is appropriate.

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Topic 2: Graphing quadratic functions.

Graph:

This quadratic function was an example used in the last topic. I won’t repeat the steps here in finding the vertex.

Both the direction and the y–intercept of a quadratic function are very easy to determine. This is an in-verted parabola, pointing downwards, due to the reflection of y = x2 indicated by the leading coefficient, – 1. The y–intercept is 5, or rather the point (0, 5), depending on how you want to call it. The y–intercept of any function, by definition, occurs on the y–axis where x = 0. If you let x = 0 in this function then you get the constant term, y = 5.



The remaining part is to plot enough points in order to draw a decent, quality graph. I tend to prefer around seven points. Here we’ll do eight points by including the y–intercept. If you are my student then I will expect you to plot the y–intercept and the vertex, in addition to another five or six points (when practical).

The secret to plotting more points is to use the proportional relationship between changes in x versus changes in y as found in the leading term, – x2. In this example, if you move a distance of 1 to either side of the vertex then y will change by – 1 since – (1)2 = – 1 due to this leading term. If you move 2 units to either side of the vertex then y will change by – 4 since – (2)2 = – 4. If you move 3 units to either side of the vertex then y will change by – 9 since – (3)2 = – 9. Once you get the hang of this process, and once you know the vertex, then plotting additional points is a matter of counting.

First I will show my final graph of this function before I mark it up.

Below I showed how I counted in order to generate new points on the graph. The fact that fractions are involved is not a problem. If you think of fractions such as 29/4 in mixed fraction form, 7 1/4, then adding and subtract-ing integers is much easier. From there you just estimate position on your graph. Suppose that you were using graph paper instead of free handing it like I did. If the squares are small enough then it would make sense to count four squares as one unit. Then one square would be one-fourth of a unit and you could produce a very nice, accurate graph.

Note also how I took advantage of the line symmetry of the parabola in graphing. Whether you move left or right of the vertex, the amount of vertical change is the same on either side.

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________________________________________

Graph:

The y–intercept is the easiest. Another reason to find the y–intercept is because it serves as a check point. If you get the wrong vertex then the y–intercept probably won’t look like it fits your graph.

Now let’s find the vertex.

Now we need to use the vertical stretch factor, 2, to generate more points for the graph. The ratios are doubled for the y–values. For example, if you move one unit to either side of the vertex then the graph moves 2(1)2 = 2 units up. If you move two units to either side of the vertex then the graph moves 2(2)2 = 8 units up. And if you move three units to either side of the vertex then the graph moves 2(3)2 = 18 units up. (That last one might be a little hard to fit on the graph.)



You always have a fall back position as far as graphing. You can always plug one or more x values into your function and calculate the corresponding y value(s). That is a mechanical process that you should want to get away from. Here you are supposed to be learning some mathematical analysis instead of doing rote calcula-tions. Nevertheless, as a trouble shooting method for a graph that isn’t looking right it’s not a bad idea at all to check yourself this way.


3. Graph:

4. Graph:

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Solutions to Reinforcement Problems

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203Section 4.3 — Quadratic Functions and Their Properties

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Section 4.4 — Quadratic Models

We shall now exploit the properties of quadratic functions towards the goal of finding the maximum or minimum of quadratic models.

Topic 1: Quadratic function models.Topic 2: The Laffer Curve.

Practice Problems: 3 – 21 odd

Topic 1: Quadratic function models.

In section 3.6 we discussed creating functions to describe mathematical models. We also mentioned that a higher purpose for doing so is the ability to optimize the function, which means to find it’s maximum or minimum points. Most optimization problems require techniques of calculus, but not models using quadratic functions. The vertex of a parabola, as we’ve seen, is the maximum or minimum of a the function, depending on which direction it is pointed.

Our first example revisits an example I worked in section 3.6, but in the context of optimization.

Suppose that you have 1,000 feet of fencing and want to construct a rectangular enclosure with four enclosures in this configuration.

What dimensions produce the maximum area, and what is that area?

The first order of business is to produce the function describing area in terms of one of the variables. I won’t go through all the steps again, but this is how we labeled the diagram and choose our variables.

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And here is the function for area in terms of width.

This function is quadratic. If you multiply it out you have the following.

This is a downward pointing parabola due to the negative leading coefficient. Therefore the vertex of this pa-rabola provides the maximum we are looking for.

It is easy to verify that the length will be 250 feet. Thus we can write the answer.

I would like to show you an alternative way of getting the vertex of a parabola whenever the quadratic func-tion is written in factored form. I find this to be very fast. Sometimes I do these in my head with confidence.

Let’s take the quadratic function for area and find the roots. We do so by setting it equal to zero and solv-ing.

If we multiply both sides by 2 then we eliminate the fraction on the left side.

We solve by factoring by setting each factor equal to 0.

So the roots are the following.

The graph of the parabola for area looks like this.



What I wanted to point out is that the horizontal coordinate of the vertex of a quadratic function is always the midpoint of its roots. So if you end up with a function in factored form, which often happens in this section, some of you may find it easier to work out the vertex this way,________________________________________A Norman window has the shape of a rectangle with a semicircle on top. Suppose that this window is to have an outer perimeter of 24 feet. What is the maximum amount of area this window can cover?

This problem was given as reinforcement problem 2 in section 3.6 as a math modeling exercise. There is a similar one in your textbook in section 4.1 as an optimization problem. I am assuming that you worked the function describing area earlier, but if not, you ought to.

The diagram used in the solution to the earlier problem is shown on the next page. Here is the function for area in terms of the radius of the circular part.

This, too, is a quadratic function, but a little messy to handle. Let’s factor the r2 term out and write this in a more traditional quadratic form, with the r2 term first followed by the r term.

Now we can find the r–coordinate of the vertex by using the little formula approach.

Let’s not be heroic this time and try to get an exact answer using all the symbols. The radius that maxi-mizes the area of the window is about 3.36 feet. I will use all the decimals in my calculator and plug this into

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the quadratic function for area in order to get the most accurate, approximated answer.

________________________________________The price p and the demand x of a certain product obey this demand equation.

x = – 30p + 1800, 0 ≤ p ≤ 60What price should be charged in order to maximize revenue?

The demand is how economists measure how many of an item can be sold (x) based upon the price that is charged (p). Simple models like this assume a linear (straight line) relationship between price and demand. This means that the assumption is that a steady rise in price will cause a steady decrease in demand, and vice versa, at the same constant rate.

In this example, charging a price of $0, i.e., giving the product away, means that you can “sell” 1,800 items. At the other extreme, charging $60 (of higher) will drop the demand to zero. The price would be too high and no one would buy any.

The wonderful aspect about capitalism is that when a free market is in place prices tend to settle in at a level that supports the demand. If someone tries to charge too much then they will suffer a loss in sales due to competition and end up making less profit.

Revenue is another word for gross sales. Suppose that this company wants to charge $40 for each item. According to demand, x = – 30p + 1800, they will sell 600 items. The revenue would be the total sales dollars earned from selling 600 items at $40 each, or $24,000.

Revenue, then, equals the price times the number of items sold, or price times demand.

By substitution we can create the revenue function in terms of price only,

Well, what do you know? This is a quadratic function, a downward pointing parabola. The roots of this function are p = 0 and p = 60. This means that the best price to charge in order to maximize revenue is the middle of 0 and 60, namely $30. Here is the picture.

(You can verify by other means that p = 30 is the horizontal coordinate for the vertex if you’d like.)Let’s write the answer to the problem.



As a matter of fact, the maximum revenue in this model is $27,000. As you can see, raising the price by $10 means they earn $3,000 less.

The fact that 30 appeared in the demand equation is a coincidence, by the way.


1. A farmer has 1,200 feet of fencing and would like to build a rectangular cattle pen alongside the river, which has a straight bank. He does not need to put any fence along the river. What dimensions will give him the largest area, and how much area will that be?

2. A hotel with 200 rooms finds that it can fill every room by charging $36. Every time they raise the price by a dollar, two more rooms go vacant. This results in the following demand equation.

x = 272 – 2p, 36 ≤ p ≤ 136What price should be charged in order to generate the most revenue? What is the maximum revenue?

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Topic 4: The Laffer Curve.

I like to end this section in regular classes with a brief discussion of the Laffer Curve. Named for Arthur Laffer, an economist at my alma mater, Univ. of Southern Calif., this is a simple demonstration of the relationship between taxes levied and government revenue. It is not a quadratic function, but it resembles an optimization problem. Laffer, himself, claims that it is not an original idea, but actually dates back to an Islamic scholar, Ibn Khaldūn (1332 – 1406). The controversy over the Laffer Curve is mostly political. The point of this coarse economic model is to demonstrate that there is a level of over taxation whereby lowering taxes will actually bring in more revenue for the government. No one knows what that point is, however, so the Laffer Curve is of no practical use since we don’t know exactly what it looks like.

The Laffer Curve is constructed by first considering two extremes. Suppose that the government collected no taxes, a tax rate of 0%. How much revenue would the government take in from taxes? None, of course. Now consider the other extreme. If the government collects all income, a tax rate of 100%, how much revenue would the government take in? Assuming that people are not made to work as slaves, there would be no revenue from taxation because who in their right mind would work if everything was kept by the government?

In between the extremes of 0% and 100% taxation some people will work because they get to keep some-thing for themselves. This means that the government would collect a positive amount of revenue. The curve is continuous; there are no breaks in it. This means that you could connect the x–intercepts with a curve meandering through the first quadrant, something like this.

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Again I must stress that no one knows where the peak of this curve lies. If maximizing government revenue is the goal, and many will argue about whether that really is a good thing, then an economy on the left side will gain more government revenue by raising tax rates, and an economy on the right side will gain more government revenue by cutting tax rates. In my opinion, people who believe we are on the left side and thus are under taxed tend to be politicians, socialists, and people who wouldn’t pay any taxes anyway. But a lot of politics enters into the perception of whether we are over taxed or under taxed. Many people believe they are over taxed but that the rich are under taxed. At some unknown point the rich are over taxed as well, and quit engaging in activity that would generate more government revenue.

President Kennedy understood these concepts intuitively in 1962 and his tax cut ideas were implemented in 1964. President Reagan was greatly influenced by the Laffer Curve and succeeded in passing a significant tax cut in 1981. President G. W. Bush’s tax cuts have arguably stimulated the economy and also have led to more government revenue in spite of the damage the economy suffered as a result of the 9/11 terrorist attacks.

It is not my purpose to persuade you that our society lives on the right or left side of the Laffer Curve, but there is evidence from these tax cuts that the country was over taxed at the time according to the Laffer Curve because the result was economic growth and growth of government revenue. You don’t have to look far to find people who will argue about all this.

My purpose here is in bringing to you a simple mathematical idea related to optimization, and a way of looking at public tax policy. Just like in the price–demand example in the last topic, there is a point at which greed becomes counterproductive. The price of anything obviously has an influence on our decisions or behavior, and taxation is a price of government.

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Solutions to Reinforcement Problems

chapter 4 — linear and quadratic functionsnofrillspublishing.com/math1141/book/5.pdf · 2009. 12....

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