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MATH 30 WEEK 7 SESSION 7 LESSON MATH 30 C1 SESSION 7 LESSON 1 SESSION 7 LESSON READ: Chapter 3, Sections 1, 2, 3 KEY POINTS Section 3.1 Most textbooks give an overview of two forms of the quadratic function: what’s known as the standard form, written as 𝑓 𝑥 = 𝑎(𝑥 − ℎ)) + 𝑘 [page 331] and the general form 𝑓 𝑥 = 𝑎𝑥) + 𝑏𝑥 + 𝑐 . I prefer the general form since the coefficients a, b, and c are required for the quadratic formula. We’ll ignore the standard form and work only with the general form. i. Quadratic function of the general form: 𝑓 𝑥 = 𝑎𝑥) + 𝑏𝑥 + 𝑐. Page 330. Starting with the function, there are a number of things that you must be able to do, which are all covered in this section: • Find the y-intercept: evaluate the function at zero. Notice that there’s an easy shortcut for this, because 𝑓 0 = 𝑐. The coordinates of the y-intercept are 0, 𝑐 • Find the x-intercepts: set f(x) equal to zero and solve for x. This is the same as solving a quadratic equation; refer to Section 1.5 for review. MATH 30 C1 INSTRUCTOR: PIRADEE NGANRUNGRUANG SESSION 7 LESSON 2 • Find the vertex of the parabola. Page 334. The vertex is the lowest point on the graph if the parabola opens upward, and the highest point on the graph if the parabola opens downward. Remember the vertex is a point on the graph—an ordered pair (x,y). Find the value of x first: 𝑥 = 12 )3 . Then plug the x-value into the function to determine the y-value. • Ascertain if the vertex is a minimum or a maximum of the function. This is the same as ascertaining if the parabola opens up or down. It is determined by the sign of the leading coefficient, a. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Page 330. • You do not need to draw graphs but you should be able to determine and interpret relevant points [intercepts, vertex] from graphs. ii. Strategy for solving problems involving maximizing or minimizing quadratic functions. Page 340. These are called “optimization” problems and there are many examples. As you’ve seen from the chapter illustrations, there are quadratic applications for any sport where an object is struck or thrown into the air. There are also important business applications; the profit function, for example, is often a quadratic model. Practice Exercises: 5, 8, 15, 16, 40, 42, 65, 67, 73 MATH 30 C1 INSTRUCTOR: PIRADEE NGANRUNGRUANG SESSION 7 LESSON 3 Section 3.2 i. Definition of a polynomial. Page 348. The degree of a polynomial is a non-negative integer. [Does “nonnegative” mean that the degree of a polynomial can be zero? Yes. Any constant, say, 5, is a polynomial of degree zero— because it can be written as 5𝑥5.] And note that a linear function is a 1st-degree polynomial, while a quadratic function is a 2nd-degree polynomial. But 𝑓 𝑥 = 𝑥6 ) is not a polynomial, because the power is not an integer, and 𝑓 𝑥 = 𝑥17 is not a polynomial, because the power is negative. [Don’t be fooled by 𝑥 = 6 89 , which is the same function as 𝑓 𝑥 = 𝑥17.] ii. Smooth, continuous graphs. Page 349. But sometimes, when modeling data, we don’t bother smoothing the curves—see practice problem #76. iii. End behavior and the Leading Coefficient Test. Pages 349- 351. End behavior refers to the behavior of the graph as x approaches infinity and negative infinity. The leading coefficient test can be applied in two ways. If the function is known, the end behavior of its graph can be determined. And if we have only the graph of a function, we can determine if the degree is odd or even, and if the leading coefficient is positive or negative. iv. Zeroes of polynomial functions. Pages 352-353. The zeroes of a function are horizontal intercepts—by definition these are the points where f(x) = 0. MATH 30 C1 INSTRUCTOR: PIRADEE NGANRUNGRUANG SESSION 7 LESSON 4 v. Finding zeroes of a polynomial function. Pages 352-353. The approach is the same as with any function: set f(x) = 0 and solve for x. For 2nd-degree (quadratic) functions, we know that all of them can be solving using the quadratic formula. However, for polynomials of degree 3 and higher, there is no equivalent to the quadratic formula. These polynomials can’t be solved algebraically unless they can be simplified by factoring into polynomials of lower degree— such is the case with Examples 5 and 6 on pages 352-353. Example 5 is a 3rd-degree polynomial and Example 6 is a 4thdegree polynomial, but by factoring, they are both reduced to products of lower-degree functions. Go over the examples to be sure the strategy is clear. vi. Multiplicities of zeroes. Page 354. This simply refers to the fact that many polynomials have two or more real zeroes at a single value of x. A function such as 𝑓 𝑥 = 𝑥 − 1 7 has three factors, as is evident if we write the function this way: 𝑓 𝑥 = 𝑥 − 1 𝑥 − 1 𝑥 − 1 . The three real zeroes of the function are: x = 1, x = 1, x = 1. vii. Intermediate value theorem. Page 355. This is common sense; we can’t get from a positive y-value to a negative yvalue, or vice versa, without crossing the x-axis. viii. Turning points. Page 356. A polynomial of degree n has at most n - 1 turning points. Notice the turning points represent local minimums and maximums. Practice Exercises: 1-18, 20, 22, 24, 26, 29, 30, 34, 40, 76 Section 3.3 i. Long division of polynomials. Pages 364-368. MATH 30 C1 INSTRUCTOR: PIRADEE NGANRUNGRUANG SESSION 7 LESSON 5 In Section 2.6 we performed arithmetic with functions, using the same methods as we use for arithmetic with real numbers. Now, since we can divide a number by a non-zero number using long division, we can do something very similar with polynomial functions. The idea is that since we can always divide an integer by a lesser integer, so we can divide a polynomial of degree n by a polynomial of lesser degree. We use the same terminology as the integer case: dividend, divisor, quotient, and remainder. You can refresh your memory of these terms by reviewing the division algorithm, below. ii. Division algorithm. Page 367. This looks complicated but is completely straightforward—don’t let the functional notation throw you. Imagine you are dividing with integers: 100/30 = 3 with a remainder of 10. 100 is the dividend, 30 is the divisor, 3 is the quotient, and 10 is the remainder. Plug the integers into the division algorithm: 100 = 30 times 3 + 10. iii. Procedure for long division of polynomials. Page 366. Step 4. is the place where the most errors occur. You must subtract the product. Remember that two negatives make a positive: 𝑥) − −4𝑥) = 𝑥) + 4𝑥) = 5𝑥) . Work through Examples 1, 2, and 3 to get comfortable with the procedure. iv. Synthetic division. Pages 368-370. BE CAREFUL! Both methods for polynomial division require you to write the dividend in descending order of powers. If MATH 30 C1 INSTRUCTOR: PIRADEE NGANRUNGRUANG SESSION 7 LESSON 6 a power of x is missing, it must be inserted with a coefficient of zero. For example: if the dividend is 6𝑥> + 5𝑥7 + 3𝑥 − 5, notice that there is no term with a power of 2. Rewrite as: 6𝑥> + 5𝑥7 + 0𝑥) + 3𝑥 − 5 v. Remainder theorem. Page 371. vi. Factor theorem. Pages 371-372. The factor theorem is a neat illustration of polynomial structure. Let’s write a polynomial first in standard form and then in factored form: 𝑓 𝑥 = 2𝑥7 − 3𝑥) − 11𝑥 + 6 𝑓 𝑥 = (𝑥 − 3)(2𝑥 − 1)(𝑥 + 2) What does the factor (x – 3) signify? That x = 3 is a zero of the function—a point where f(x) = 0. Check: 𝑓 3 = 54 − 27 − 33 + 6 = 0 Practice Exercises: 4, 6, 11, 14, 15, 43, 44, 51