1 / 99

9.1 Quadratic Functions

9.1 Quadratic Functions. Section 9.1 p 1. Example 1. Unlike linear functions, quadratic functions can be used to describe situations where the speed or rate of change of an object changes as it moves.

travis-albert
Télécharger la présentation

9.1 Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 9.1 Quadratic Functions Section 9.1 p1

  2. Example 1 Unlike linear functions, quadratic functions can be used to describe situations where the speed or rate of change of an object changes as it moves. The height in feet of a ball thrown upward from the top of a building after t seconds is given by h(t) = −16t2 + 32t + 128, t ≥ 0. Find the ball’s height after 0, 1, 2, 3, and 4 seconds and describe the path of the ball. Solution We have h(0) = −16 · 02 + 32 · 0 + 128 = 128 ft. h(1) = −16 · 12 + 32 · 1 + 128 = 144 ft. h(2) = −16 · 22 + 32 · 2 + 128 = 128 ft. Continuing, we get Table 9.1. The ball rises from 128 ft when t = 0 to a height of 144 ft when t = 1, returns to 128 ft when t = 2 and reaches the ground, height 0 ft, when t = 4. Table 9.1Height of a ball Figure 9.1: Height of a ball thrown from the top of a building Section 9.1 p2

  3. Interpreting Quadratic Functions Expressed in Standard Form The expression defining the function in Example 1 has a term, −16t2, involving the square of the independent variable. This term is called the quadratic term, the expression is called a quadratic expression, and the function it defines is called a quadratic function. In general, A quadratic function is one that can be written in the form y = f(x) = ax2 + bx + c, a, b, c constants, a ≠ 0. • The expression ax2 + bx + c is a quadratic expression in standard form. • The term ax2 is called the quadratic term or leading term, and its coefficient a is the leading coefficient. • The term bx is called the linear term. • The term c is called the constant term. Section 9.1 p3

  4. Example 2 Interpreting Quadratic Functions Expressed in Standard Form (continued) For the function h(t) = −16t2 + 32t + 128 in Example 1, interpret in terms of the ball’s motion (a) The constant term 128 (b) The sign of the quadratic term −16t2. Solution (a) The constant term is the value of the function when t = 0, and so it represents the height of the building. (b) The negative quadratic term −16t2 counteracts the positive terms 32t and 128 for t > 0, and eventually causes the values of h(t) to decrease, which makes sense since the ball eventually starts to fall to the ground. Section 9.1 p4

  5. Interpreting Quadratic Functions Expressed in Standard Form (continued) Figure 9.1 shows an important feature of quadratic functions. Unlike linear functions, quadratic functions have graphs that bend. This is the result of the presence of the quadratic term. See Figure 9.2, which illustrates this for the power functions f(x) = x2 and g(x) = −x2. The shape of the graph of a quadratic function is called a parabola. If the coefficient of the quadratic term is positive then the parabola opens upward, and if the coefficient is negative the parabola opens downward. Figure 9.2: Graphs of f(x) = x2 and g(x) = −x2 Section 9.1 p5

  6. Example 3 Solution When t = 4, the factor (t − 4) has the value 4 − 4 = 0. So h(4) = −16(0)(4 + 2) = 0. In practical terms, this means that the ball hits the ground 4 seconds after it is thrown. When t = −2, the factor (t + 2) has the value −2 + 2 = 0. However, there is no practical interpretation for this, since the domain of h is t ≥ 0. See Figure 9.3. (See next slide for Figure 9.3.) Interpreting Quadratic Functions Expressed in Factored Form Factoring a quadratic expression puts it in a form where we can easily see what values of the variable make it equal to zero. The function h(t) = −16t2 + 32t + 128 in Example 1 can be expressed in the form h(t) = −16(t − 4)(t + 2), t ≥ 0. What is the practical interpretation of the factors (t − 4) and (t + 2)? Section 9.1 p6

  7. Example 3 (continued) Interpreting Quadratic Functions Expressed in Factored Form (continued) (Continued) Solution (continued) (Continued) Figure 9.3: Interpretation of the factored form h(t) = −16(t − 4)(t + 2) Section 9.1 p7

  8. Interpreting Quadratic Functions Expressed in Factored Form (continued) Values of the independent variable where a function has the value zero, such as the t = 4 and t = −20 in the previous example, are called zeros of the function. In general, we have the following definition. A quadratic function in x is expressed in factored form if it is written as y = f(x) = a(x − r)(x − s), where a, r, and s are constants and a ≠ 0. • The constants r and s are zeros of the function f(x) = a(x − r)(x − s). • The constant a is the leading coefficient, the same as the constant a in the standard form. The zeros of a function can be easily recognized from its graph as the points where the graph crosses the horizontal axis. Section 9.1 p8

  9. Can each function graphed in Figure 9.4 be expressed in factored form f(x) = a(x − r)(x − s)? If so, is each of the parameters r and s positive, negative, or zero? (Assume r ≤ s.) Figure 9.4 Example 4abc Interpreting Quadratic Functions Expressed in Factored Form (continued) Solution Since r and s are the x-intercepts, we look at whether the x-intercepts are positive, negative or zero. (a) There is only one x-intercept at x = 0, so r = s = 0. (b) There are two x-intercepts, one positive and one negative, so r < 0 and s > 0. (c) There is only one x-intercept. It is positive and r = s. Section 9.1 p9

  10. Can each function graphed in Figure 9.4 be expressed in factored form f(x) = a(x − r)(x − s)? If so, is each of the parameters r and s positive, negative, or zero? (Assume r ≤ s.) Figure 9.4 Example 4de Interpreting Quadratic Functions Expressed in Factored Form (continued) Solution Since r and s are the x-intercepts, we look at whether the x-intercepts are positive, negative or zero. (d) There are no x-intercepts, so it is not possible to write the quadratic in factored form. (e) There are no x-intercepts, so it is not possible to write the quadratic in factored form. Section 9.1 p10

  11. Example 5 Interpreting Quadratic Functions Expressed in Factored Form (continued) Expressing a quadratic function in factored form allows us to see not only where it is zero, but also where it is positive and where it is negative. A college bookstore finds that if it charges p dollars for a T-shirt, it sells 1000 − 20p T-shirts. Its revenue is the product of the price and the number of T-shirts it sells. (a) Express its revenue R(p) as a quadratic function of the price p in factored form. (b) For what prices is the revenue positive? Solution (a) The revenue R(p) at price p is given by Revenue = R(p) = (Price)(Number sold) = p(1000 − 20p). (b) We know the price, p, is positive, so to make R positive we need to make the factor 1000−20p positive as well. Writing it in the form 1000 − 20p = 20(50 − p) we see that R is positive only when p < 50. Therefore the revenue is positive when 0 < p < 50. Section 9.1 p11

  12. The function h(t) = −16t2 + 32t + 128 in Example 1 can be expressed in the form h(t) = −16(t − 1)2 + 144. Use this form to show that the ball reaches its maximum height h = 144 when t = 1. Example 6 Solution Looking at the right-hand side, we see that the term −16(t − 1)2 is a negative number times a square, so it is always negative or zero, and it is zero when t = 1. Therefore h(t) is always less than or equal to 144 and is equal to 144 when t = 1. This means the maximum height the ball reaches is 144 feet, and it reaches that height after 1 second. See Table 9.2 and Figure 9.5. (See next slide for Table 9.2 and Figure 9.5.) Interpreting Quadratic Functions Expressed in Vertex Form The next example illustrates a form for expressing a quadratic function that shows conveniently where the function reaches its maximum value. Section 9.1 p12

  13. (See previous slide.) Example 6 (continued) Interpreting Quadratic Functions Expressed in Vertex Form (continued) Solution (continued) (Continued) Table 9.2Values of h(t) = −16(t − 1)2 + 144 are less than or equal to 144 Figure 9.5: Ball reaches its greatest height of 144 ft at t = 1 Section 9.1 p13

  14. Interpreting Quadratic Functions Expressed in Vertex Form (continued) The point (1, 144) in Figure 9.5 is called the vertex of the graph. For quadratic functions the vertex shows where the function reaches either its largest value, called the maximum, or its smallest value, called the minimum. In Example 6 the function reaches its maximum value at the vertex because the coefficient is negative in the term −16(t − 1)2. In general: A quadratic function in x is expressed in vertex form if it is written as y = f(x) = a(x − h)2 + k, where a, h, and k are constants and a ≠ 0. For the function f(x) = a(x − h)2 + k, • f(h) = k, and the point (h, k) is the vertex of the graph. • The coefficient a is the leading coefficient, the same a as in the standard form. ・ If a > 0 then k is the minimum value of the function, and the graph opens upward. ・ If a < 0 then k is the maximum value of the function, and the graph opens downward Section 9.1 p14

  15. For each function, find the maximum or minimum and sketch the graph, indicating the vertex. (a) g(x) = (x − 3)2 + 2 Example 7a Interpreting Quadratic Functions Expressed in Vertex Form (continued) Solution (a) The expression for g is in vertex form. We have g(x) = (x − 3)2 + 2 = Positive number (or zero) + 2. Thus g(x) ≥ 2 for all values of x except x = 3, where it equals 2. The minimum value is 2, and the vertex is at (3, 2) where the graph reaches its lowest point. See Figure 9.6(a). Figure 9.6: Interpretation of the vertex form of a quadratic expression Section 9.1 p15

  16. For each function, find the maximum or minimum and sketch the graph, indicating the vertex. (b) A(t) = 5 − (t + 2)2 Example 7b Interpreting Quadratic Functions Expressed in Vertex Form (continued) Solution (b) The expression for A is also in vertex form. We have A(t) = 5 − (t + 2)2 = 5 − Positive number (or zero). Thus A(t) ≤ 5 for all t except t = −2, where it equals 5. The maximum value is 5, and the vertex is at (−2, 5), where the graph reaches its highest point. See Figure 9.6(b). Figure 9.6: Interpretation of the vertex form of a quadratic expression Section 9.1 p16

  17. For each function, find the maximum or minimum and sketch the graph, indicating the vertex. (c) h(x) = x2 − 4x + 4 Example 7c Interpreting Quadratic Functions Expressed in Vertex Form (continued) Solution (c) The expression for h is not in vertex form. However, recognizing that it is a perfect square, we can write it as h(x) = x2 − 4x + 4 = (x − 2)2, which is in vertex form with k = 0. So h(2) = 0 and h(x) is positive for all other values of x. Thus the minimum value is 0, and it occurs at x = 2. The vertex is at (2, 0). See Figure 9.6(c). Figure 9.6: Interpretation of the vertex form of a quadratic expression Section 9.1 p17

  18. The functions in Example 7 ((a) g(x) = (x − 3)2 + 2, (b) A(t) = 5 − (t + 2)2 and (c) h(x) = x2 − 4x + 4) can be thought of as resulting from shifts or scales of the basic function f(x) = x2. Describe these operations in each case. Example 8 Interpreting Quadratic Functions Expressed in Vertex Form (continued) Solution (a) The function g(x) is formed from f(x) by a horizontal shift 3 units to the right and a vertical shift up 2 units. (b) The function A(x) is formed from f(x) by a horizontal shift 2 units to the left followed by a vertical shift up 5 units and a multiplication by −1. (c) Using the form h(x) = (x − 2)2 shows that h(x) is formed from f(x) by a horizontal shift 2 units to the right. Section 9.1 p18

  19. 9.1 QUADRATIC FUNCTIONS Key Points • Expressing quadratic functions in different forms reveals different properties of the function • The standard form shows the vertical intercept • The factored form shows where the function is equal to zero • The vertex form shows where the function reaches its maximum or minimum Section 9.1 p19

  20. 9.2Working With Quadratic Expressions Section 9.2 p20

  21. In the previous section we saw how to interpret the form in which a quadratic function is expressed. In this section we see how to construct and manipulate expressions for quadratic functions.. Section 9.2 p21

  22. Constructing Quadratic Expressions In the previous section we saw that different forms give us different information about a quadratic function. The standard form tells us the vertical intercept, the factored form tells us the horizontal intercepts, and the vertex form tells us the maximum or minimum value of the function. We can also go the other way and use this information to construct an expression for a given quadratic function. Section 9.2 p22

  23. Find a quadratic function whose graph could be Figure 9.7: Find possible expressions for these functions Example 1a Constructing Quadratic Expressions (continued) Solution (a) Since we know the zeros, we start with a function in factored form: f(x) = a(x − r)(x − s). The graph has x-intercepts at x = 1 and x = 4, so the function has zeros at those values, so we choose r = 1 and s = 4, which gives f(x) = a(x − 1)(x − 4). Since the y-intercept is −12, we know that y = −12 when x = 0. So −12 = a(0 − 1)(0 − 4) −12 = 4a −3 = a. So the function f(x) = −3(x − 1)(x − 4) has the right graph. Notice that the value of a is negative, which we expect because the graph opens downward. Section 9.2 p23

  24. Find a quadratic function whose graph could be Figure 9.7: Find possible expressions for these functions Example 1b Constructing Quadratic Expressions (continued) Solution (b) We are given the vertex of the parabola, so we try to write its equation using the vertex form y = f(x) = a(x − h)2 + k. Since the coordinates of the vertex are (3,−13), we let h = 3 and k = −13. This gives f(x) = a(x − 3)2 − 13. The y-intercept is (0, 5), so we know that y = 5 when x = 0. Substituting, we get 5 = a(0 − 3)2 − 13 5 = 9a − 13 18 = 9a 2 = a. Therefore, the function f(x) = 2(x − 3)2 − 13 has the correct graph. Notice that the value of a is positive, which we expect because the graph opens upward. Section 9.2 p24

  25. Example 2ab Constructing Quadratic Expressions (continued) In the next example we see how a quadratic function can be used to represent profit. The revenue to a bookstore from selling 1000 − 20p T-shirts at p dollars each is R(p) = p(1000 − 20p). Suppose that each T-shirt costs the bookstore $3 to make. (a) Write an expression for the cost of making the T-shirts. (b) Write an expression for the profit, which is the revenue minus the cost. Solution (a) Since each T-shirt costs $3, we have Cost = 3(Number of T-shirts sold) = 3(1000 − 2p). (b) We have Profit = Revenue − Cost = p(1000 − 20p) − 3(1000 − 20p). Section 9.2 p25

  26. The revenue to a bookstore from selling 1000 − 20p T-shirts at p dollars each is R(p) = p(1000 − 20p). Suppose that each T-shirt costs the bookstore $3 to make. (c) For what values of p is the profit positive? Example 2c Constructing Quadratic Expressions (continued) Solution (c) Factored form is the most useful for answering this question. Taking out a common factor of (1000 − 20p), we get Profit = p(1000 − 20p) − 3(1000 − 20p) = (p − 3)(1000 − 20p) = 20(p − 3)(50 − p). This first factor is p − 3, which is positive when p > 3 and negative when p < 3. So p > 50 : g(p) = 20(p − 3)(50 − p) = positive × negative = negative 3 < p < 50 : g(p) = 20(p − 3)(50 − p) = positive × positive = positive p < 3 : g(p) = 20(p − 3)(50 − p) = negative × positive = negative. So the profit is positive if the price is greater than $3 but less than $50. Section 9.2 p26

  27. Converting Quadratic Expressions to Standard and Factored Form In the previous section we saw three forms for a function giving the height of a ball: h(t) = −16t2 + 32t + 128 (standard form) = −16(t − 4)(t + 2) (factored form) = −16(t − 1)2 + 144 (vertex form). One way to see that these forms are equivalent is to convert them all to standard form. Section 9.2 p27

  28. Converting Quadratic Expressions to Standard and Factored Form (continued) Converting to Standard Form We convert an expression to standard form by expanding and collecting like terms, using the distributive law. For example, to check that −16(t− 1)2 + 144 and −16t2 + 32t+ 128 are equivalent expressions, we expand the first term: −16(t − 1)2 + 144 = −16(t2 − 2t + 1) + 144 = −16t2 + 32t − 16 + 144 = −16t2 + 32t + 128. Similarly, expanding shows that −16(t − 4)(t + 2) and −16t2 + 32t + 128 are equivalent: −16(t − 4)(t + 2) = −16(t2 − 4t + 2t − 8) = −16(t2 − 2t − 8) = −16t2 + 32t + 128. Section 9.2 p28

  29. Example 3a Write each of the following expressions in the indicated form. (a) 2 − x2 + 3x(2 − x) (standard) Converting Quadratic Expressions to Standard and Factored Form (continued) Converting to Factored Form Factoring takes an expression from standard form to factored form, using the distributive law in reverse. Solution (a) Expanding and collecting like terms, we have 2 − x2 + 3x(2 − x) = 2 − x2 + 6x − 3x2 = −4x2 + 6x + 2. Section 9.2 p29

  30. Example 3b Write each of the following expressions in the indicated form. Converting Quadratic Expressions to Standard and Factored Form (continued) Converting to Factored Form Solution (b) This is already almost in factored form. All we need to do is express the division by 2 as multiplication by 1/2: Section 9.2 p30

  31. Example 3c Write each of the following expressions in the indicated form. (c) (z + 3)(z − 2) + z − 2 (factored) Converting Quadratic Expressions to Standard and Factored Form (continued) Converting to Factored Form Solution (c) We could expand this expression and then factor it, but notice that there is a common factor of z − 2, which enables us to get to the factored form more directly: (z + 3)(z − 2) + z − 2 = (z − 2)((z + 3) + 1) factor out (z − 2) = (z − 2)(z + 4) = (z − 2)(z − (−4)). Section 9.2 p31

  32. Example 3d Write each of the following expressions in the indicated form. (d) 5(x2 − 2x + 1) − 9 (vertex) Converting Quadratic Expressions to Standard and Factored Form (continued) Converting to Factored Form Solution (d) The key to putting this in vertex form is to recognize that the expression in parentheses is a perfect square: In Example 3(d) we had to rely on recognizing a perfect square to put the expression in vertex form. Now we give a more systematic method. Section 9.2 p32

  33. Example 4 How Do We Put an Expression in Vertex Form? Find the vertex of the parabolas (a) y = x2 + 6x + 9 (b) y = x2 + 6x + 8. Solution (a) We recognize the expression on the right-hand side of the equal sign as a perfect square: y = (x + 3)2, so the vertex is at (−3, 0). (b) Unlike part (a), the expression on the right-hand side of the equal sign is not a perfect square. In order to have a perfect square, the 6x term should be followed by a 9.We can make this happen by adding a 9 and then subtracting it in order to keep the expressions equivalent: add and subtract 9 = (x + 3)2 − 1 since −9 + 8 = −1. Therefore, y = x2 + 6x + 8 = (x + 3)2 − 1. This means the vertex is at the point (−3,−1). Section 9.2 p33

  34. How Do We Put an Expression in Vertex Form? (continued) In the last example, we put the equation y = x2 + 6x + 8 into a form where the right-hand side contains a perfect square, in a process that is called completing the square. To complete the square, we use the form of a perfect square (x + p)2 = x2 + 2px + p2. Section 9.2 p34

  35. Put each expression in vertex form by completing the square. (a) x2 − 8x Example 5a How Do We Put an Expression in Vertex Form? (continued) Solution (a) We compare the expression with the form of a perfect square: x2 − 8x x2 + 2px + p2. To match the pattern, we must have 2p = −8, so p = −4. Thus, if we add (−4)2 = 16 to x2 − 8x we obtain a perfect square. Of course, adding a constant changes the value of the expression, so we must subtract the constant as well. add and subtract 16 x2 − 8x = (x − 4)2 − 16. Notice that the constant, 16, that was added and subtracted could have been obtained by taking half the coefficient of the x-term, (−8/2), and squaring this result. This gives (−8/2)2 = 16. Section 9.2 p35

  36. Put each expression in vertex form by completing the square. (b) x2 + 4x − 7 Example 5b How Do We Put an Expression in Vertex Form? (continued) Solution (b) We compare with the form of a perfect square: x2 + 4x − 7 x2 + 2px + p2. Note that in each of the previous examples, we chose p to be half the coefficient of x. In this case, half the coefficient of x is 2, so we add and subtract 22 = 4: add and subtract 4 x2 + 4x − 7 = (x + 2)2 − 11. Section 9.2 p36

  37. Put the expression on the right-hand side in vertex form by completing the square. (a) y = x2 + x + 1 Example 6a An Alternative Method for Completing the Square Solution (a) The overall strategy is to add or subtract constants from both sides to get the right-hand side in the form of a perfect square. We first subtract 1 from both sides so that the right-hand side has no constant term, then add a constant to complete the square: y = x2 + x + 1 y − 1 = x2 + xsubtract 1 from both sides y − 1 + ¼ = x2 + x + ¼ add 1/4 = (1/2)2 to both sides y − ¾ = (x + ½)2 Section 9.2 p37

  38. Put the expression on the right-hand side in vertex form by completing the square. (b) y = 3x2 + 24x − 15 Example 6b An Alternative Method for Completing the Square (continued) Solution (b) This time we first divide both sides by 3 to make the coefficient of x2 equal to 1, then proceed as before: y = 3x2 + 24x − 15 y/3 = x2 + 8x − 5 divide both sides by 3 y/3 + 5 = x2 + 8xadd 5 to both sides so there is no constant term on the right y/3 + 5 + 42 = x2 + 8x + 42add 42 = (8/2)2 to both sides y/3 + 21 = (x + 4)2 y/3 = (x + 4)2 − 21 y = 3(x + 4)2 − 63 isolate y on the left side. Section 9.2 p38

  39. A bookstore finds that if it charges $p for a T-shirt then its revenue from T-shirt sales is given by R = f(p) = p(1000 − 20p) What price should it charge in order to maximize the revenue? Example 7 An Alternative Method for Completing the Square (continued) Solution We first expand the revenue function into standard form: R = f(p) = p(1000 − 20p) = 1000p − 20p2. Since the coefficient of p2 is −20, we know the graph of the quadratic opens downward. So the vertex form of the equation gives us its maximum value. The expression on the right is more complicated than the ones we have dealt with so far, because it has a coefficient −20 on the quadratic term. In order to deal with this, we work with the equation R = 1000p − 20p2 rather than the expression 1000p − 20p2: (Solution continued on next slide.) Section 9.2 p39

  40. A bookstore finds that if it charges $p for a T-shirt then its revenue from T-shirt sales is given by R = f(p) = p(1000 − 20p) What price should it charge in order to maximize the revenue? Example 7 (continued) An Alternative Method for Completing the Square (continued) Solution (continued) (continued from previous slide) R = 1000p − 20p2 −R/20 = p2 − 50p divide by −20 −R/20 + 252 = p2 − 50p + 252 add 252 to both sides −R/20 + 252 = (p − 25)2 R = −20(p − 25)2 + 20 · 252 R = −20(p − 25)2 + 12,500. So the vertex is (25, 12,500), indicating that the price that maximizes the revenue is p = $25. Figure 9.8 shows the graph of f. The graph reveals that revenue initially rises as the price increases, but eventually starts to fall again when the high price begins to deter customers. (See next slide for Figure 9.8.) Section 9.2 p40

  41. (See previous slide.) Example 7 (continued) An Alternative Method for Completing the Square (continued) Solution (continued) (continued) Figure 9.8: Revenue from the sale of T-shirts Notice two differences between the method in Example 7 and the method in Examples 4 and 5: we can move the −20 out of the way temporarily by dividing both sides, and instead of adding and subtracting 252 to an expression, we add to both sides of an equation. Since in the end we subtract the 252 from both sides and multiply both sides by −20, we end up with an equivalent expression on the right-hand side. Section 9.2 p41

  42. Visualizing The Process of Completing The Square We can visualize how to find the constant that needs to be added to x2 + bx in order to obtain a perfect square by thinking of x2 + bx as the area of a rectangle. For example, the rectangle in Figure 9.9 has area x(x + 8) = x2 + 8x. Figure 9.9: Rectangle with sides x and x + 8 Now imagine cutting the rectangle into pieces as in Figure 9.10 and trying to rearrange them to make a square, as in Figure 9.11. The corner piece, whose area is 42 = 16, is missing. By adding this piece to our expression, we “complete” the square: x2 + 8x + 16 = (x + 4)2. Figure 9.10: Cutting off a Figure 9.11: Rearranging the strip of width 4 piece tomake a square with a missing corner Section 9.2 p42

  43. 9.2 WORKING WITH QUADRATIC EXPRESSIONS Key Points • Converting between different forms of quadratic expressions • Completing the square Section 9.2 p43

  44. 9.3Solving Quadratic Equations by Completing the Square Section 9.3 p44

  45. When we find the zeros of a quadratic function f(x) = ax2 + bx + c, we are solving the equation ax2 + bx + c = 0. A quadratic equation in x is one which can be put into the standard form ax2 + bx + c = 0, where a, b, c are constants, with a ≠ 0. Section 9.3 p45

  46. Example 1 Some quadratic equations can be solved by taking square roots. Solve (a) x2 − 4 = 0 (b) x2 − 5 = 0 Solution (a) Rewriting the equation as x2 = 4, we see the solutions are (b) Similarly, the solutions to x2 − 5 = 0 are Section 9.3 p46

  47. Example 2 Use the result of Example 1 to solve (x − 2)2 = 5. Solution Since the equation x2 = 5 has solutions the equation (x − 2)2 = 5 has solutions Thus the solutions to (x − 2)2 = 5 are which can be combined as Using a calculator, these solutions are approximately x = 2 + 2.236 = 4.236 and x = 2 − 2.236 = −0.236. Section 9.3 p47

  48. Example 3 Solve (a) 2(y + 1)2 = 0 (b) 2(y − 3)2 + 4 = 0. Solution (a) Since 2(y + 1)2 = 0, dividing by 2 gives (y + 1)2 = 0 so y + 1 = 0. Thus the only solution is y = −1. It is possible for a quadratic equation to have only one solution. (b) Since 2(y − 3)2 + 4 = 0, we have 2(y − 3)2 = −4, so dividing by 2 gives (y − 3)2 = −2. But since no number squared is −2, this equation has no real number solutions. Section 9.3 p48

  49. In general, if we can put a quadratic equation in the form (x − h)2 = Constant, then we can solve it by taking square roots of both sides. Section 9.3 p49

  50. Example 4 For the function h(t) = −16(t − 1)2 + 144 giving the height of a ball after t seconds, find the times where the ball reaches a height of 135 feet. Solution We want to find the values of t such that h(t) = 135, so we want to solve the equation −16(t − 1)2 + 144 = 135. Isolating the (t − 1)2 term we get Therefore, the solutions are t = 0.25 and t = 1.75. So the ball reaches a height of 135 ft on its way up very soon after being thrown and again on its way down about 2 seconds after being thrown. Next we develop a systematic method for solving quadratic equations by taking square roots. The solutions to quadratic equations are sometimes called roots of the equation. Section 9.3 p50

More Related