Quadratic
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The document discusses quadratic functions. It defines quadratic functions as functions of the form f(x) = ax^2 + bx + c where a ≠ 0. It explains that the graph of a quadratic function is a parabola. If a>0, the parabola opens up and there is a minimum point. If a<0, the parabola opens down and there is a maximum point. The roots of the quadratic function correspond to the x-intercepts of the graph. If b^2 - 4ac > 0 there are two distinct roots, if b^2 - 4ac = 0 there is one repeated root, and if b^2 - 4ac < 0 there are no real roots
Quadratic
- 1. QUADRATIC FUNCTIONS The word quadratic comes from the Latin word Quadratus which means square
- 2. Chapter Objectives Understand the concept of quadratic functions and their graphs. Find maximum and minimum values of quadratic Functions. Sketch graphs of quadratic functions Understand and use the concept of quadratic Inequalities.
- 3. Recognising quadratic functions f : x a x 2 + b x + c f (x) = a , b and c are constants a 0 The highest power of x is 2 x 2 x c + + a b
- 4. Determine whether each of the following is a quadratic function. f(x) = 2x 2 h(x) = 4x - 3x + 1 2 g(x) = x + 3x 2 k(x) = 5x + 7 g(x) = 3x + 1 2 x h(x) = (x + 3) + 4 2
- 5. a 0 f(x) = ax + bx + c 2 quadratic function 2 ax + bx + c quadratic expression 2 ax + bx + c = 3 quadratic equation
- 6. Plotting the graphs of quadratic functions Based on given tabulated values By constructing the table of values
- 7. x f ( x ) -4 -3 0 -2 -1 1 2 3 4 -7 9 8 0 5 The table below shows some values of x and the corresponding values of f(x) of the function f(x) = 9 – x 2 8 5 0 -7 Plot the graph of the function Example 0 -2 -1 1 2 5 Select suitable scales on both axes and subsequently plot the Graph. Given the quadratic function f (x) = x 2 – 2x – 4. Plot the graph of the function for -3 ≤ x ≤ 5. We first construct the table of values of the function. 3 4 -5 -4 4 -1 -4 -1 4 11 11 -3 x f ( x )
- 8. Shapes of graphs of quadratic functions f(x) = ax 2 + bx + c If a > 0 , then the graph of the function is a parabola with a min pt. If a < 0 , then the graph of the function is a parabola with a max pt. a > 0 a < 0 axes of symmetry Minimum point Maximum point
- 9. Example Describe the shape of the graph of each of the following quadratic functions. Solution (a) f (x) = - 3x 2 – 4x + 5 (b) g (x) = 10x 2 + 6x + 3 (a) Since a = - 3 < 0 , the graph of the function is a parabola with a maximum pt. (b) Since a = 10 > 0 , the graph of the function is a parabola with a minimum pt.
- 10. Relating the position of the graph of a quadratic function f(x) = ax 2 + bx + c with the for types of roots f(x) = 0 m n Referring to the graph, When f(x) = 0 , x = m and x = n m and n are the roots of the equation m and n are also the values of x where the graph intersects the x – axis. Therefore , the roots of f(x) = ax 2 + bx + c are the points where the graph of f(x) intersects the x – axis . Values of x when f(x) = 0
- 11. In this respect , we have three cases : (I) If f (x) = ax 2 + bx + c has two distinct (different) roots , meaning b 2 – 4ac > 0 , then the graph of the function f (x) intersects at two distinct points. x x a > 0 a < 0
- 12. Example 1 f (x) = 2x 2 –x -10 Points of intersection with the x – axis. f(x) = 0 When f(x) = 0 b 2 – 4ac = (-1) 2 – 4(2)(-10) = 81 , b 2 – 4ac > 0 a > 0 Hence 2x 2 –x -10 = 0 (2x – 5)( x + 2) = 0 X = 5/2 , -2
- 13. Example 2 f (x) = -x 2 + 3x +10 Points of intersection with the x – axis. f(x) = 0 When f(x) = 0 b 2 – 4ac = 3 2 – 4(-1)(10) = 49 , b 2 – 4ac > 0 a < 0 Hence -x 2 + 3x +10 = 0 (5 - x)( x + 2) = 0 X = 5 , -2
- 14. In this respect , we have three cases : (II) If f (x) = ax 2 + bx + c has two real and equal roots , meaning b 2 – 4ac = 0 , then the graph of the function f (x) intersects at only one point. x x a > 0 a < 0
- 15. Example 3 f (x) = x 2 +6x + 9 When f(x) = 0 b 2 – 4ac = 6 2 – 4(1)(9) = 0 , b 2 – 4ac = 0 a > 0 Hence Point of intersection with the x – axis. f(x) = 0 x 2 + 6x + 9 = 0 (x + 3)( x + 3) = 0 x = -3
- 16. In this respect , we have three cases : (III) If f (x) = ax 2 + bx + c does not have any real roots , meaning b 2 – 4ac < 0 , then the graph of the function f (x) does not intersect the x - axis. x x a > 0 a < 0
- 17. Example 4 f (x) = 2x 2 + 5x + 7 There is NO point of intersection with the x – axis. When f(x) = 0 b 2 – 4ac = (5) 2 – 4(2)(7) = - 31 , b 2 – 4ac < 0 a > 0 Hence
- 18. Summary (I) If b 2 – 4ac > 0 x x a < 0 (II) If b 2 – 4ac = 0 x x a < 0 a > 0 a > 0 (III) If b 2 – 4ac < 0 x x a < 0 a > 0