Week 10. Topic 1 Polynomial Functions

Transcription

1 Week 10 Topic 1 Polnomial Functions 1 Week 10 Topic 1 Polnomial Functions Reading Polnomial functions result from adding power functions 1 together. Their graphs can be ver complicated, so the come up in a variet of applications where the simpler graphs of our basic functions aren t quite right. We alread use some polnomial functions: f() = 2 and g() = 3 are ver simple polnomials, and we ve seen more complicated cubics like h() = ( 1)( + 2) 2. In fact, here s the graph of h(): 5-5 While we re looking at the graph, let s look at some relationships between the graph and the equation. If we epand h(), we get h() = Power functions are in the form f() = k n, so like 3 2 or The eponents are non-negative integers onl.

2 Week 10 Topic 1 Polnomial Functions 2 You ma remember from Math 65 that the polnomial has degree 3. In general, the degree is the biggest eponent 2. Degree of a Polnomial Function Degree tells us a few things about the graph: For a polnomial function with Degree = n 1. The graph has at most n zeros For eample, a quadratic is degree 2 and has 0, 1, or 2 zeros. 2. The graph turns around at most (n 1) times For eample, a quadratic turns around once; the cubic above turns around 2 times. The function f() = 3 never turns around. 3. In the long-run, the graph looks like the graph of the term with the highest degree To see what point 3 means, let s look at the cubic equation h() = at different zoom levels: See how the dip becomes less and less noticeable as we zoom out further and further? Think about what happens when ou put a big number into h(). 2 Assuming the eponents are all positive and there s onl one variable

3 Week 10 Topic 1 Polnomial Functions 3 For eample, h(10) = The first term is 1000 and the second term is onl 300. Or evaluate h(100) = The first term is 1,000,000, and the second term is onl 30,000. The 3 term grows so much faster than the 2 term (even compared with 3 2 ), that eventuall it dominates what the graph looks like. The other terms contributions to the total output (-value) are smaller and smaller as gets bigger. Zeroes of a Polnomial Function Degree gives us some information. What else can we get? If h() = ( 1)( + 2) 2, here s the graph again: 5-5 The zeros of h are -2 and 1. Notice the formula for h has factors of ( 1) and ( + 2). B the zero-product principle, when ou substitute in -2, ( + 2) = 0, which makes h = 0. If ou have a polnomial in factored form, identifing the zeros is prett eas 3, but we can see even more than that. Notice we could also write h() = ( 1)( + 2)( + 2). The ( + 2) is a repeated factor. We sa it has multiplicit 2. The factor ( 1) onl appears once. It has multiplicit 1. 3 And if ou have a function like g() = ( + 1)( + 2)( + 3), the onl zeros it can have are 0, -1, -2, and -3. Don t add others to our graph.

4 Week 10 Topic 1 Polnomial Functions 4 Look at the graph s behavior near = -2. See how it touches the -ais but turns around without passing through? Not so at = 1 where the graph goes right through the ais without a pause. In general, an time a polnomial has a zero of even multiplicit (2, 4, 6, etc.), the graph touches at that point but does not pass through. The function will pass through the -ais for an zero with odd multiplicit (1, 3, 5, etc.) This makes sense when ou look at the graph of a quadratic function with onl one real zero. The graph of f() = ( + 2) 2 looks like this: This is the graph of = 2 shifted left 2 units. It has the same behavior near = -2 as the graph of h() on the previous page: it touches the ais and turns around. Of course, the -values are negative for h() and positive for f(); the other factors affect what direction it goes. 5 Eample 1: Sketch the graph of P() = ( 1) 2 (3 + 2)( + 3) 3 Solution: First, get the zeroes and the multiplicit: Zero Multiplicit Behavior 1 2 Touches and turns around, like a parabola -2/3 1 Passes right though, like a line -3 3 Flattens out and passes through, like = 3 Another eas point to get is the -intercept: P(0) = (0 1) 2 ( )(0 + 3) 3 P(0) = 54

5 Week 10 Topic 1 Polnomial Functions 5 Now put those on the graph: Known Points We also need the function s long-run behavior. If we epand out the polnomial, we get: P() = The long run behavior is like = 3 6, so we know P() is decreasing, has some zeroes and ups and downs, and then increases again eventuall. Notice we don t even need to epand the whole polnomial, we can just look at the pieces with in them and their coefficients. (Look at the formula on the previous page, make sure that makes sense.) We ll add long-run behavior arrows to our graph: Known Points <-- Long-Run Behavior Arrow Long-Run Behavior Arrow --> Now we pla Graph Detective and fill in the missing parts of the curve. Look again at the table with the multiplicit of each zero. As we read from left to right, the graph decreases, flattens out near (-3, 0), and passes through. The graph has to pass through (-2/3, 0), so it has to turn around (we don t know where eactl, but we know it has to) and go through there.

6 Week 10 Topic 1 Polnomial Functions 6 After that, it has to hit the -ais at (1, 0) and turn around. So it must increase to some maimum near = 0 (again, not sure where eactl, could be on either side) pass through (0, 54), and decrease to (1, 0). Then it can hit the ais, turn around, and meet up with the long-run behavior arrow. Here s what the actual graph looks like. P() We need a graphing utilit to get this eactl, but we can figure out the basic shape and get a rough idea with just a little information from the equation. As long as our graph shows all that information (intercepts, multiplicit, long-run behavior, detective work in-between), ou re doing great! Eample 2: Find a formula for the polnomial function shown in this graph. Note the point (0, 9) is on the graph. Q() Solution: Looking at the long-run behavior, we know it s a polnomial of odd degree, with an - ais reflection. Let s make a table to summarize the information from the zeroes:

7 Week 10 Topic 1 Polnomial Functions 7 Zero Multiplicit Factor of Q() = -4 2 (or 4, 6, 8, etc.) ( + 4) 2 = 2 3 (or 5, 7, 9, etc.) ( 2) 3 Based on this information, we can write the function as: Q() = a( + 4) 2 ( 2) 3 We put the constant a out front as a shape factor. We need that to make Q(0) = 9 and account for the -ais reflection. We solve for a using the -intercept: And our final function is to be sure matches the graph. 9 = a(0 + 4) (0 3) 9 = a = a Q( ) ( 4) ( 2) = +, which we check on the calculator Summar In conclusion, if ou want to graph a polnomial function Identif the zeros and the behavior near the zeros Identif the long-run behavior Sketch a smooth continuous curve that fits the conditions above Eamples 1 and 2 are ver common problems. If the final eam has a big question about polnomials (and it probabl will), it ll look ver similar to one of those two. Rational functions net week keep all the same ideas we have for polnomials, so get these down and it ll be smoother sailing for the rest of the term. Tetbook Book pages (ou can skip the last bit about building cubic models from data). The section seems long, but there are onl a few ke ideas. You have to know all of them to sketch a good graph of a polnomial though, so take notes or make flash cards or whatever ou need to do so that ou can see a polnomial and instantl think of all the things ou know.

8 Week 10 Topic 1 Polnomial Functions 8 Be especiall careful not to mi up our facts about polnomials with facts about eponentials. The both have formulas with eponents, but polnomial and power functions have numbers in the eponents. Eponential functions have the variable in the eponent.