1. 3 3.1Introduction to Functions 3.2Quadratic Functions 3.3Graphs of Other Functions Chapter Summary Case Study Functions and Graphs (1)

2. P. 2 Therefore, if the height of the building is 80 m, while the speed of the lift is 2 m/s, then the time it takes to reach the top floor from the ground floor is (80  2) s = 40 s. In junior forms, we learnt that the relationship between time, speed and distance can be expressed as: Time  Distance Speed The above relationship can be treated as a function and we will discuss the concept of a function in details in Section 3.1 of this chapter. Case Study How long does it take for us to reach the top floor of the building by taking this lift? It depends on the speed of the lift and the height of the building.

3. P. 3 A. Basic Idea of a Function The concept of a function is one of the basic concepts in mathematics. Suppose the price of petrol is $12 per litre. The following table shows amount paid and the corresponding volume of petrol bought. Volume of petrol bought (L)510152025 Amount paid ($)60120180240300 If we use $y to represent the amount paid and x L to represent the volume of petrol bought, then we have: y  12x The amount paid$12 multiplied by the volume of petrol boughtis 3.1 Introduction to Functions

4. P. 4 A. Basic Idea of a Function The value of y (the amount paid) depends on a given value of x (the volume of petrol bought). We call y the dependent variable and x the independent variable. Each value of x gives one and exactly one value of y. A variable y is said to be a function of a variable x if there is a relation between x and y such that every value of x gives exactly one value of y. In this case, we say that y is a function of x. y  12x The amount paid$12 multiplied by the volume of petrol boughtis 3.1 Introduction to Functions

5. P. 5 B. Different Ways to Represent Functions Volume of petrol bought (L)0510152025 Amount paid ($)060120180240300  Tabular representation  Algebraic representation y  12x  Graphical representation Plot the corresponding values of x and y according to the above table. The relation between x and y can be expressed in different ways: 3.1 Introduction to Functions

6. P. 6 B. Different Ways to Represent Functions Consider a square with side x cm and perimeter y cm. x (cm)12345 y (cm)48121620  Tabular representation  Algebraic representation y  4x  Graphical representation Plot the corresponding values of x and y according to the above table. We can use the graph to find the value of y for other value of x, for example: When x  2.5, y  10. 3.1 Introduction to Functions

7. P. 7 A function can be expressed by an equation, such as y  3x + 2. It is common to use the notation f(x) to denote a function of x, such as f(x)  3x + 2.  The symbol ‘f(x)’ is read as ‘f of x’, Notes: y  3x + 2 and f(x)  3x + 2 represent the same function.  f(x) does not mean f is multiplied by x.  f(x) represents the value of the function at x. For example, f(10) is the value of the function f(x) at x  10, that is, f(10)  3(10) + 2  32 Consider the representations f(x)  3x + 2 and y  3x + 2. The meaning of f(10) is the same as that of the phrase ‘the value of y when x  10’.  The letter ‘f ’ in the notation f(x) may be replaced by other letters to represent different functions of x, such as g(x), h(x), F(x) and G(x). C. Notation for a Function 3.1 Introduction to Functions

8. P. 8 Example 3.1T Consider f(x)  x 3 – 4x 2 + 5. (a)Find f(–2), f(0) and f(2). (b)Is f(–2) + f(2)  f(0)? Solution: (a)f(–2)  (–2) 3 – 4(–2) 2 + 5  –19 f(0)  (0) 3 – 4(0) 2 + 5  5 f(2)  (2) 3 – 4(2) 2 + 5  –3 (b)f(–2) + f(2)  –19 + (–3)  –22 ∴ f(–2) + f(2)  f(0) In general, f(a + b)  f(a) + f(b). and f(0)  5 C. Notation for a Function 3.1 Introduction to Functions

9. P. 9 Example 3.2T Consider f(x)  k – 3x 2 and f(3)  –7. (a)Find the value of k. (b) Find the values of x such that f(x)  8. Solution: (a) f(3)  –7 k  20 (b)From (a), f(x)  20 – 3x 2. ∴ 20 – 3x 2  8 Substitute x  3 into f(x), then find the value of k by solving the equation f(3)   7. k – 3(3) 2  –7 k – 27  –7 ∵ f(x)  8 3x 2  12 x 2  4 x   2 C. Notation for a Function 3.1 Introduction to Functions

10. P. 10 Example 3.3T Consider f(x)  –4x 2 + 7x – 2. (a)Express f(–a) and f(–a + 1) in terms of a. (b)If f(–a)  f(–a + 1), find the value of a. Solution: (a) f(x)  –4x 2 + 7x – 2 f(–a)  –4(–a) 2 + 7(–a) – 2 (b) ∵ f(–a)  f(–a + 1) ∴ –4a 2 – 7a – 2  –4a 2 + a + 1  –4a 2 – 7a – 2 f(–a + 1)  –4(–a + 1) 2 + 7(–a + 1) – 2  –4(a 2 – 2a + 1) – 7a + 7 – 2  –4a 2 + 8a – 4 – 7a + 7 – 2  –4a 2 + a + 1 –8a  3 a  C. Notation for a Function 3.1 Introduction to Functions

11. P. 11 D. Domain and Co-domain of a Function When describing a function y  f(x), we may want to know: (i)Domain (ii)Co-domain (iii)Range all possible values that the independent variable x can take Examples:  Consider f(x)  x 2 + 1. x can take any real numbers. ∴ The domain is the set of all real numbers.  Consider g(x) . x cannot be 0. ∴ The domain is the set of all real numbers except 0. The domain of the function y  is the set of all real numbers except 1. 3.1 Introduction to Functions

12. P. 12 D. Domain and Co-domain of a Function (i)Domain: the set of all real numbers (iii)Range: the set of real numbers larger than or equal to 1 (ii)Co-domain: the set of all real numbers For the function y  x 2 + 1, even though x can be negative, y must be positive. When describing a function y  f(x), we may want to know: (i)Domain (ii)Co-domain (iii)Range all possible values that the independent variable x can take all possible values that the dependent variable y can take all output values of dependent variable y of the function, that is, the corresponding values of independent variable x e.g.Consider f(x)  y  x 2 + 1. 3.1 Introduction to Functions

13. P. 13 E. Variables in Functions We have been using x and y as the independent and the dependent variables of a function. In fact, we can have functions of other variables:  Consider f(l)  l 2. f(l) is a function of l.  Consider h(  )  sin . h(  ) is a function of .  Consider g(t) . g(t) is a function of t. We can also use different variables for the same function. For example, f(l)  l 2, f(x)  x 2 and f(t)  t 2 represent the same function. We call the variables l, x and t the dummy variables. 3.1 Introduction to Functions

14. P. 14 A. Graphs of Quadratic Functions 3.2 Quadratic Functions A function in the form y  ax 2 + bx + c, where a, b and c are constants and a  0 is called a quadratic function. Plot the graph of y  –x 2 – 2x + 6 for –4  x  2: x–4–3–2–1012 y–236763 The graph is a curve which is called a parabola. y   x 2  2x  6

15. P. 15 A. Graphs of Quadratic Functions Example 3.4T Solution: Consider y  x 2 + 2x – 1. (a)Complete the following table. (b)Plot the graph of y  x 2 + 2x – 1 for –4  x  2. x–4–3–2–1012 y x–4–3–2–1012 y72 –2–127 (a) y  x 2 + 2x – 1 (b) Refer to the figure on the right. y  x 2  2x  1 3.2 Quadratic Functions

16. P. 16 B. Properties of Quadratic Functions The following figure shows two parabolas y  ax 2 + bx + c where a > 0 and a < 0 respectively.  Each parabola has a vertex.  Each parabola is symmetrical about the axis of symmetry.  The parabolas cut the y-axis at (0, c) and c is called the y-intercept. 3.2 Quadratic Functions

17. P. 17 1.Direction of opening: (a)If a > 0, then it opens upwards. (b)If a < 0, then it opens downwards. Properties of the graph of a quadratic function y  ax 2 + bx + c: 2.Since the graph cuts the y-axis at (0, c), c is the y-intercept of the graph. 3.(a)If a > 0, then the vertex is the lowest point of the parabola. (b)If a < 0, then the vertex is the highest point of the parabola. 4.The axis of symmetry passes through the vertex. B. Properties of Quadratic Functions 3.2 Quadratic Functions

18. P. 18 Apart from determining the direction of opening of the graph, the values of a (in y  ax 2 + bx + c) also affect the width of the opening. When a > 0, the opening is narrower for a larger value of a. When a < 0, the opening is wider for a larger value of a. (i.e., the more negative the value of a, the narrower the opening.) B. Properties of Quadratic Functions 3.2 Quadratic Functions

19. P. 19 Example 3.5T Consider the graphs of the following three functions: (I)y  (x – 3) 2 – 1 (II) y  2(x – 3) 2 – 1 (III)y  –(x – 3) 2 + 1 (a)Compare the shapes of the graphs of (I) and (III). (b)Describe the difference in the shapes of the graphs between (I) and (II). Solution: (a)Consider function (III): y  –(x – 3) 2 + 1  –[(x – 3) 2 – 1] ∴ (I) and (III) are symmetrical about the x-axis. (b)The coefficients of the x 2 terms in (I) and (II) are positive. ∴ The opening of the graph of (I) is wider than that of (II). The coefficient of the x 2 term in (II) is greater than that in (I). B. Properties of Quadratic Functions 3.2 Quadratic Functions

20. P. 20 C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions Consider the graph of y  ax 2 + bx + c: Suppose the graph cuts the x-axis at two points (p, 0) and (q, 0). The two values p and q are called the x-intercepts of the graph of y  ax 2 + bx + c. Since the y-coordinates of these two points are 0, ax 2 + bx + c  0. ∴ The x-intercepts of the graph of y  ax 2 + bx + c satisfy the equation ax 2 + bx + c  0. ∴ The x-intercepts of the graph of y  ax 2 + bx + c are the roots of the quadratic equation ax 2 + bx + c  0.

21. P. 21 Example 3.6T The figure shows the graph of y  5x 2 + 8x – 4. Solve 5x 2 + 8x – 4  0 graphically. Solution: The graph cuts the x-axis at the points (–2.0, 0) and (0.4, 0). (cor. to the nearest 0.05) Hence the roots of 5x 2 + 8x – 4  0 are –2.0 and 0.4. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

22. P. 22 Example 3.7T The figure shows the graph of y  4x 2 + 12x + 9. Solve 4x 2 + 12x + 9  0 graphically. Solution: The graph touches the x-axis at the point (–1.5, 0). (cor. to the nearest 0.05) Hence the root of 4x 2 + 12x + 9  0 is –1.5. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

23. P. 23 Example 3.8T The figure shows the graph of y  –x 2 + x – 1. Solve –x 2 + x – 1  0 graphically. Solution: The graph does not intersect with the x-axis. Hence the equation –x 2 + x – 1  0 has no real roots. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

24. P. 24 Example 3.9T Consider the graph of y  x 2 + 3x + 1. (a)Solve x 2 + 3x + 1  0 graphically. (b)Solve the above equation by the quadratic formula and compare the result with (a). Solution: (a)The graph cuts the x-axis at the points (–2.6, 0) and (–0.4, 0). Hence the roots of x 2 + 3x + 1  0 are –2.6 and –0.4. (cor. to 1 d. p.) (b)By the quadratic formula, we have According to the scale of the graph, we can only find the approximate roots of the equation correct to 1 decimal place in (a) while the roots in (b) are exact values. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

25. P. 25 D. Quadratic Graphs and Nature of Roots In Chapter 2, we learnt that the discriminant  of a quadratic equation can help us to determine the nature of the roots. We can also use   b 2 – 4ac to find the number of x-intercepts of the graph of y  ax 2 + bx + c. If   b 2 – 4ac > 0, then the graph of y  ax 2 + bx + c cuts the x-axis at two distinct points. If   b 2 – 4ac  0, then the graph of y  ax 2 + bx + c touches the x-axis at only one point. If   b 2 – 4ac < 0, then the graph of y  ax 2 + bx + c does not cut the x-axis. 3.2 Quadratic Functions

26. P. 26 D. Quadratic Graphs and Nature of Roots   b 2 – 4ac Nature of roots of the equation ax 2 + bx + c  0 Number of x-intercepts of the graph of y  ax 2 + bx + c  > 0 Two unequal real roots2   0 One double real root (Two equal real roots) 1  < 0 No real roots0 The following table summarizes the nature of roots of quadratic equations and the corresponding number of x-intercepts of their graphs. 3.2 Quadratic Functions

27. P. 27 Example 3.10T The graph of y  (2 – k)x 2 – 16x – 16 touches the x-axis at only one point. (a)Find the value of k. (b)Hence solve (2 – k)x 2 – 16x – 16  0. Solution: (a)Since the graph touches the x-axis,   0. (b)The equation is –4x 2 – 16x – 16  0. ∴ x 2 + 4x + 4  0 16 2 – 4(2 – k)(–16)  0 256 + 128 – 64k  0 64k  384 k  6 (x + 2) 2  0 x  –2 D. Quadratic Graphs and Nature of Roots 3.2 Quadratic Functions

28. P. 28 A. Linear Functions 3.3 Graphs of Other Functions A function in the form y  mx + c, where m and c are constants and m  0 is called a linear function.  If m  0, then the function becomes y  c which is called a constant function. Notes:  c is the y-intercept of the graph. If c  0, then the graph is a straight line passing through the origin. m  0m  0

29. P. 29 B. Other Functions 3.3 Graphs of Other Functions Apart from linear and quadratic functions, Example 3.11T and Example 3.12T show some other functions.

30. P. 30 The function y  ax 3 + bx 2 + cx + d, where a, b, c and d are constants and a  0, is called a cubic function. Example 3.11T Consider the function y  x 3 – 2x 2 – x + 2. (a)Plot the graph of y  x 3 – 2x 2 – x + 2 for –2  x  3. (b)Find the x-intercepts and the y-intercept of the graph.(Give the answers correct to 1 decimal place if necessary.) Solution: x–2–1.5–1–0.50 y–12–4.401.92 x0.511.522.53 y1.10–0.602.68 (a) (b)The x-intercepts are –1.0, 1.0 and 2.0. The y-intercept is 2.0. B. Other Functions 3.3 Graphs of Other Functions

31. P. 31 Example 3.12T The figure shows the graphs of y  3(2 – x), y  x 2 + 9 and y  – 2. Compare the graphs of the functions with respect to the following: (a)domain (b)existence of maximum or minimum value (c)number of x-intercepts Solution: (a)For y  3(2 – x), the domain is the set of all real numbers. For y  – 2, the domain is the set of all real numbers except 0. For y  x 2 + 9, the domain is the set of all real numbers. B. Other Functions 3.3 Graphs of Other Functions

32. P. 32 Example 3.12T The figure shows the graphs of y  3(2 – x), y  x 2 + 9 and y  – 2. Compare the graphs of the functions with respect to the following: (a)domain (b)existence of maximum or minimum value (c)number of x-intercepts Solution: (b)For y  3(2 – x), there is no maximum or minimum value. For y  – 2, there is no maximum or minimum value. For y  x 2 + 9, there is a minimum value. (c)For y  3(2 – x), there is one x-intercept. For y  – 2, there is one x-intercept. For y  x 2 + 9, there is no x-intercept. B. Other Functions 3.3 Graphs of Other Functions

33. P. 33 3.1 Introduction to Functions Chapter Summary 1.A variable y is said to be a function of x if there is a relation between x and y such that every value of x corresponds to exactly one value of y. 2.We use f(x) to denote a function of x, where x is the independent variable. 3.The domain is the set of values that the independent variable can take. The co-domain is the set of possible values that the dependent variable can take. The range is the set of values that the dependent variable takes corresponds to the independent variable.

34. P. 34 3.2 Quadratic Functions Chapter Summary Properties of the graph of y  ax 2 + bx + c: 1.(a)If a > 0, then the graph opens upwards. (b)If a < 0, then the graph opens downwards. 2.The graph cuts the y-axis at (0, c), that is, the y-intercept  c. 3.The vertex is the turning point of the graph. (a)If a > 0, the vertex is the minimum point of the graph. (b)If a < 0, the vertex is the maximum point of the graph. 4.The axis of symmetry is a line that the graph is symmetrical about it. It passes through the vertex of the graph.

35. P. 35 3.2 Quadratic Functions Chapter Summary For two quadratic functions: (I)y  a 1 x 2 + b 1 x + c 1 (II)y  a 2 x 2 + b 2 x + c 2 If 0 a 1 > a 2, then graph (II) opens narrower than the graph (I). The roots of the quadratic equation ax 2 + bx + c  0 (a  0) can be obtained by finding the x-intercept(s) of the graph.   b 2 – 4ac Number of real roots of the equation ax 2 + bx + c  0 Number of x-intercepts of the graph of y  ax 2 + bx + c  > 0 22   0 11  < 0 00

36. P. 36 3.3 Graphs of Other Functions Chapter Summary 1.The graph of a linear function y  mx + c is a straight line. 2.Without drawing the graph of a function y  f(x), the y-intercept can be found by putting x  0 into the function, that is, the y-intercept of y  f(x) is f(0).

37. Follow-up 3.1 Solution: C. Notation for a Function 3.1 Introduction to Functions Consider g(x)  (x – 1)(x + 2). (a)Find g(3), g(4), g(9) and g(12). (b)Determine whether each of the following is true. (i)g(3)  g(4)  g(12)(ii) [g(3)] 2  g(3 2 ) (a)g(3)  (3 – 1)(3 + 2)  10 g(4)  (4 – 1)(4 + 2)  18 g(9)  (9 – 1)(9 + 2)  88 (b)(i)g(3)  g(4)  10  18  180 ∴ g(3)  g(4)  g(12) g(12)  (12 – 1)(12 + 2)  154 (ii)[g(3)] 2  10 2  100 ∴ [g(3)] 2  g(3 2 ) and g(12)  154 and g(3 2 )  g(9)  88

38. Follow-up 3.2 Solution: Consider f(x)  kx 2 + 5x and f(2)  2. (a)Find the value of k. (b) Find the values of x such that f(x)  3. (a) f(2)  2 k   2 (b)From (a), we have f(x)   2x 2 + 5x. ∴  2x 2 + 5x  3 Substitute x  2 into f(x), then find the value of k by solving the equation f(2)  2. k(2) 2 + 5(2)  2 4k + 10  2 ∵ f(x)  3 2x 2  5x + 3  0 (2x – 3)(x – 1)  0 x  1 or C. Notation for a Function 3.1 Introduction to Functions

39. Follow-up 3.3 Solution: Consider f(x)  x 2 + 2x – 3. (a)Express f(2a) and f(a – 1) in terms of a. (b)If f(2a)  4f(a – 1), find the value of a. (a) f(x)  x 2 + 2x – 3 f(2a)  (2a) 2 + 2(2a) – 3 (b) ∵ f(2a)  4f(a – 1) ∴ 4a 2 + 4a – 3  4(a 2 – 4)  4a 2 + 4a – 3 f(a – 1)  (a – 1) 2 + 2(a – 1) – 3  a 2 – 2a + 1 + 2a – 2 – 3  a 2 – 4 4a 2 + 4a – 3  4a 2 – 16 a  4a  –13 C. Notation for a Function 3.1 Introduction to Functions

40. Follow-up 3.4 Solution: Consider y  x 2 – 2x + 2. (a)Complete the following table. (b)Plot the graph of y  x 2 – 2x + 2 for –2  x  4. x–2–101234 y x–2–101234 y1052125 (a) y  x 2 – 2x + 2 (b) Refer to the figure on the right. y  x 2  2x  2 A. Graphs of Quadratic Functions 3.2 Quadratic Functions

41. Follow-up 3.5 Solution: Consider the following two functions: (I)y  16 – (x – 1) 2 (II) y  16 – 4(x – 1) 2 (a)Find the y-intercepts of the graphs of the two functions. (b)Without drawing the graphs, do you know which is opening wider? (a)y  16 – (x – 1) 2  16 – (x 2 – 2x + 1) ∴ The y-intercept of (I) is 15. (b)Since –4 < –1 < 0, the opening of the graph of (I) is wider than that of (II). y  16 – 4(x – 1) 2  16 – (4x 2 – 8x + 4) ∴ The y-intercept of (II) is 12.  –x 2 + 2x + 15  –4x 2 + 8x + 12 B. Properties of Quadratic Functions 3.2 Quadratic Functions

42. Follow-up 3.6 Solution: The figure shows the graph of y  –2x 2 + x + 3. Solve –2x 2 + x + 3  0 graphically. The graph cuts the x-axis at the points (–1.0, 0) and (1.5, 0). (cor. to the nearest 0.05) Hence the roots of –2x 2 + x + 3  0 are –1.0 and 1.5. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

43. Follow-up 3.7 Solution: The figure shows the graph of y  x 2 + 4x + 4. Solve x 2 + 4x + 4  0 graphically. Hence the root of x 2 + 4x + 4  0 is –2.0. The graph touches the x-axis at the point (–2.0, 0). (cor. to the nearest 0.05) C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

44. Follow-up 3.8 Solution: The figure shows the graph of y  –2x 2 – 1. Solve –2x 2 – 1  0 graphically. The graph does not intersect with the x-axis. Hence the equation –2x 2 – 1  0 has no real roots. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

45. Follow-up 3.9 Solution: Consider the graph of y  3x 2 + x  1. (a)Solve 3x 2 + x  1  0 graphically. (b)Solve the above equation by the quadratic formula and compare the result with (a). (a)The graph cuts the x-axis at the points (–0.75, 0) and (0.45, 0). Hence the roots of 3x 2 + x  1  0 are –0.75 and 0.45. (cor. to the nearest 0.05) (b)By the quadratic formula, we have According to the scale of the graph, we can only find the approximate roots of the equation correct to the nearest 0.05 in (a) while the roots in (b) are exact values. C. Solving Quadratic Equations by the Graphical Method the Graphical Method 3.2 Quadratic Functions

46. Follow-up 3.10 Solution: The graph of y  –x 2 + 10x + k touches the x-axis at only one point. (a)Find the value of k. (b)Hence solve –x 2 + 10x + k  0. (a)Since the graph touches the x-axis,   0. (b)The equation is –x 2 + 10x – 25  0. ∴ x 2 – 10x + 25  0 10 2 – 4(–1)(k)  0 100 + 4k  0 k  –25 (x – 5) 2  0 x  5 D. Quadratic Graphs and Nature of Roots 3.2 Quadratic Functions

47. Follow-up 3.11 Solution: Consider the function y  –x 3 + x 2 – x + 1. (a)Find the corresponding values of y for x  –1, –0.5, 0, 0.5, 1, 1.5 and 2. (Give the answers correct to 2 decimal places if necessary.) (b)Plot the graph of y  –x 3 + x 2 – x + 1 for –1  x  2. (c)Find the x-intercept and the y-intercept of the graph. x–1–0.500.511.52 y41.8810.630–1.63–5 (a) (c)The x-intercept is 1.0. The y-intercept is 1.0. B. Other Functions 3.3 Graphs of Other Functions

48. Follow-up 3.12 Solution: The figure shows the graphs of y  2x 2 + 4x + 1, y  –x 3 – 1 and y . Compare the graphs of the functions with respect to the following: (a)domain (b)existence of maximum or minimum value (c)number of x-intercepts (a)For y  2x 2 + 4x + 1, the domain is the set of all real numbers. For y  –x 3 – 1, the domain is the set of all real numbers. For y , the domain is the set of all real numbers. B. Other Functions 3.3 Graphs of Other Functions

49. Follow-up 3.12 Solution: The figure shows the graphs of y  2x 2 + 4x + 1, y  –x 3 – 1 and y . Compare the graphs of the functions with respect to the following: (a)domain (b)existence of maximum or minimum value (c)number of x-intercepts (b)For y  2x 2 + 4x + 1, there is a minimum value. For y  –x 3 – 1, there is no maximum or minimum value. For y , there is no maximum or minimum value. B. Other Functions 3.3 Graphs of Other Functions

50. Follow-up 3.12 Solution: The figure shows the graphs of y  2x 2 + 4x + 1, y  –x 3 – 1 and y . Compare the graphs of the functions with respect to the following: (a)domain (b)existence of maximum or minimum value (c)number of x-intercepts (c)For y  2x 2 + 4x + 1, there are two x-intercepts. For y  –x 3 – 1, there is one x-intercept. For y , there is no x-intercept. B. Other Functions 3.3 Graphs of Other Functions