Quadratic Functions. Before we start looking at polynomials, we should know some common terminology.

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1 Quadratic Fuctios I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively i mathematical models of productio costs, cosumer demads, wildlife maagemet, biological processes, ad may other scietific studies. Usig these fuctios ad their graphs, predictios regardig future treds ca be made. Polyomial Fuctios ad Their Graphs: Before we start lookig at polyomials, we should kow some commo termiology. Defiitio: A polyomial of degree is a fuctio of the form 1 1 ( )... P x = a x + a x + + a x + a where a 0. The umbers a0, a1, a,..., a are called the coefficiets of the polyomial. The umber a 0 is the costat coefficiet or costat term. The umber a, the coefficiet of the highest power is the leadig coefficiet, ad the term ax is the leadig term. Notice that a polyomial is usually writte i descedig powers of the variable, ad the degree of a polyomial is the power of the leadig term. For istace ( ) 3 P x = 4x x + 5 is a polyomial of degree 3. Also, if a polyomial cosists of just a sigle term, such as Q x = 7x, the it is called a moomial. ( ) 4 Polyomials of degree 0 are costat fuctios ad polyomials of degree 1 are liear fuctios, whose graphs are both straight lies. Polyomials of degree are quadratic equatios, ad their graphs are parabolas. As the degree of the polyomial icreases beyod, the umber of possible shapes the graph ca be icreases. However, the graph of a polyomial fuctio is always a smooth cotiuous curve (o breaks, gaps, or sharp corers). ( ) Moomials of the form P x = x are the simplest polyomials.

2 As the figure suggest, the graph of P(x) = x has the same geeral shape as y = x whe is eve, ad the same geeral shape as y = x 3 whe is odd. However, as the degree becomes larger, the graphs become flatter aroud the origi ad steeper elsewhere. To this poit, we have had some experiece with quadratic equatios. We kow that the graph of a quadratic equatio gives us a parabola. I this sectio, we will see how quadratic equatios (ad their correspodig parabolas) ca give us some valuable iformatio i real-life situatios. If a parabola opes upward, the we kow that there is a poit at which the graph reaches its miimum value. Likewise, if we have a parabola that opes dowward, we kow that there must be a maximum value for that fuctio. These maximum ad miimum values are referred to as extreme values. There are may practical situatios, which ca be represeted by such fuctios. I these situatios, the extreme values ca ofte give us valuable pieces of iformatio. For example, the ower of a clothig store might be iterested i kowig her maximum profit or miimum cost. As review, we will look at the defiitio of a quadratic fuctio. Def: A quadratic fuctio is a fuctio f of the form = ax + bx + c where a, b, ad c are real umbers ad a 0. Stadard Form ad Completig the Square: I determiig extreme values, it is very helpful to put the quadratic fuctio ito stadard form. This ca be doe by completig the square. Now, I ca already hear the collective groa at the metio of completig the square, but it is a very ecessary skill i determiig extreme values. I am cofidet that you ca lear it well if you just take it step-by-step. First, let s take a look at stadard form.

3 Def: A quadratic fuctio ca be expressed i the stadard form = a( x h) + k by completig the square. The graph of f is a parabola with vertex (h,k); the parabola opes upward if a > 0 or dowward if a < 0. If a > 0, the the miimum value of f occurs at x = h ad this value is words, the poit with coordiates (h,k). f ( h) = k, i other If a < 0, the the maximum value of f occurs at x = h ad this value is also the poit (h,k). f ( h) = k,

4 Example 1: Sketch the graph of the fuctio = x + 10x 0 ad state the vertex. Begi by completig the square. Solutio: Comparig this with the form a = 1 b = 10 c = 0 = ax + bx + c, we ca see that Example 1 (Cotiued): Step 1: Group the x ad x terms, ad factor out ay a value. I this case, sice a = 1, we simply have to group the first two terms. = ( x + 10x) 0 Step : After groupig the first two terms ad factorig out a, cosider the coefficiet of x, i this case 10. Divide this value by ad take the square. 10 = 5 = 5 Step 3: Complete the square by addig the above value iside the paretheses. Note: Ay value that is added to oe side of the equatio must also be subtracted from that side of the equatio i order to keep from chagig the value. i.e.- add a value of zero to the equatio. Therefore, we must subtract the total value outside the paretheses. (Here is where you must be careful to pay attetio to ay commo value that has bee factored out. We will see this i the ext example.) = ( x + 10x + 5) 0 5 Step 4: Factor the portio of the equatio i paretheses ad combie ay other like terms. = ( x + 10x + 5) 0 5 = ( x + 10x + 5) 45 = ( x + 5) 45 OR = ( x ( 5)) 45 (which will be helpful, as you will see ow)

5 The fuctio should be i the form = a( x h) + k. I this case, a = 1 h = 5 k = 45 Example 1 (Cotiued): Notice: You must pay special attetio to the sig of h. Rewritig the equatio as we did above will be helpful i seeig the sig of h more clearly. Now that we have completed the square, the remaider of the solutio is very simple. By defiitio, we kow the parabola opes upward, sice a > 0, ad we have a miimum value at the poit (-5,-45) which is also the vertex of the parabola. We are simply asked to sketch the parabola, so it is ot ecessary i this case to fid multiple poits o the graph. We ca simply sketch the graph usig what we kow. Let s look at oe more example. Example : For the fuctio = x 16x + 1 (a) Express the fuctio i stadard form. (b) Sketch its graph. (c) Fid its maximum or miimum value.

6 Solutio: (a) We ca put the fuctio ito stadard form by completig the square. Step 1: Group the x ad x terms, ad factor out ay a value. = (x 16x) + 1 = ( x 8x) + 1 Example (Cotiued): Step : After groupig the first two terms ad factorig out a, cosider the coefficiet of x, i this case ( 8). Divide this value by ad take the square. 8 = ( 4) = 16 (Notice that the sig of the x coefficiet does ot really matter at this poit, because after squarig the term, you will always have a positive umber.) Step 3: Complete the square by addig the above value iside the paretheses ad subtractig the total value outside the paretheses. = ( x 8x + 16) Note: Special attetio must be give to the value that was factored out. This will affect the value which is subtracted outside the paretheses. i.e.- (16)=3 Step 4: Factor the portio of the equatio i paretheses ad combie ay other like terms. = ( x 8x + 16) = ( x 8x + 16) 0 = ( x 4) 0 So, we have the graph i stadard form. (b) To sketch the graph, let s combie the iformatio we have from the stadard form of the equatio.

7 a = h = 4 k = 0 Sice a > 0, the parabola opes upward; therefore, we have a miimum value at x = 4. This value occurs at the poit (4,-0) which is also the vertex. With this iformatio, we ca easily sketch the graph. Example (Cotiued): (c) The miimum value of the graph occurs at x = 4, ad the value there is ( 0). This is the value f ( h) = k. Maximum or Miimum Value of a Quadratic Equatio: I the examples so far, we have bee asked to graph the fuctio, etc. Sometimes is simply ecessary to kow the maximum or miimum value. I this case, you do ot eed to complete the square, because you do ot eed the equatio to be i stadard form. Def: The maximum or miimum value of a quadratic fuctio = ax + bx + c occurs at x = b a If a > 0, the the miimum value is f b a.

8 b If a < 0, the the maximum value is f. a

9 Example 3: A coffee shop o the SAC campus has a sales aalyst cosider the store s sales records for a moth. The aalyst produces a formula i order to help the store geerate more profit. He fids that if the shop cells l orders of latte i oe day, the profit (i dollars) is give by the followig fuctio 1 P ( l) = x + 3x What is the maximum profit possible for the store i oe day, ad how may lattes must they sell i order to reach this maximum? Solutio: First, it is importat to otice that they asked for a maximum profit. We kow we will have a maximum istead of a miimum, because a < 0. Usig the defiitio above, the maximum occurs at b a l = = 3 ( 1 ) (1000) = ( 1) = 3000 = 1500 This meas that i order to receive a maximum profit, the coffee shop must sell 1500 lattes per day. I order to fid out what the maximum profit actually is (i dollars), we have to evaluate the fuctio at this value, l. b 1 P = P (1500) = (1500) a = = = 450 The maximum profit would be $ (1500) 1800

Polynomial Functions and Their Graphs

Polynomial Functions and Their Graphs Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively