2. A quadratic function always contains a term in x 2. It can also contain terms in x or a constant. Here are examples of three quadratic functions: The characteristic shape of a quadratic function is called a parabola. y = x 2 y = x 2 – 3 x y = –3 x 2

3. When the coefficient of x 2 is positive the vertex is a minimum point and the graph is U-shaped. When the coefficient of x 2 is negative the vertex is a maximum point and the graph is shaped. Parabolas have a vertical axis of symmetry … …and a turning point called the vertex.

4. When a quadratic function factorizes we can use its factorized form to find where it crosses the x -axis. For example: Sketch the graph of the function y = x 2 – 2 x – 3. The function crosses the x -axis when y = 0. x 2 – 2 x – 3 = 0 ( x + 1)( x – 3) = 0 x + 1 = 0or x – 3 = 0 x = 3 The function crosses the x -axis at the points (–1, 0) and (3, 0). x = –1

5. By putting x = 0 in y = 2 x 2 – 5 x – 3 we can also find where the function crosses the y -axis. y = 2(0) 2 – 5(0) – 3 y = – 3 So the function crosses the y -axis at the point (0, –3). The quadratic function y = ax 2 + bx + c will cross the y -axis at the point (0, c ). We now know that the function y = x 2 – 2 x – 3 passes through the points (–1, 0), (3, 0) and (0, –3) and so we can place these points on our sketch. In general:

6. 0 y x (–1, 0) (3, 0) (0, –3) (1, –4) We can also use the fact that a parabola is symmetrical to find the coordinates of the vertex. The x coordinate of the vertex is half-way between –1 and 3. When x = 1, y = (1) 2 – 2(1) – 3 y = –4 So the coordinates of the vertex are (1, –4). We can now sketch the graph.

7. When a quadratic function is written in the form y = a ( x – α )( x – β ), it will cut the x -axis at the points ( α, 0) and ( β, 0). α and β are the roots of the quadratic function. When a quadratic function is written in the form y = a ( x – α )( x – β ), it will cut the x -axis at the points ( α, 0) and ( β, 0). α and β are the roots of the quadratic function. For example, write the quadratic function y = 3 x 2 + 4 x – 4 in the form y = a ( x – α )( x – β ) and hence find the roots of the function. This function can be factorized as follows, y = (3 x – 2)( x + 2) It can be written in the form y = a ( x – p )( x – q ) as Therefore, the roots are In general:

8. In general, when the quadratic function y = ax 2 + bx + c is written in vertex form as a ( x – h) 2 + k The coordinates of the vertex will be ( h, k ). The axis of symmetry will have the equation x = h. Also: If a > 0 ( h, k ) will be the minimum point. If a < 0 ( h, k ) will be the maximum point. Plotting the y -intercept, (0, c ) will allow the curve to be sketched using symmetry.

9. The coordinates of the vertex are therefore (–2, –5). The equation of the axis of symmetry is x = –2. Also, when x = 0 we have y = –1 So the curve cuts the y -axis at the point (–1, 0). Using symmetry we can now sketch the graph. y x 0 (–1, 0) x = –2 y = x 2 + 4 x – 1 (–2, –5) Sketch y = ( x + 2) 2 – 5

11. Graphs can be transformed by translating, reflecting, stretching or rotating them. The equation of the transformed graph will be related to the equation of the original graph. When investigating transformations it is most useful to express functions using function notation. For example, suppose we wish to investigate transformations of the function f ( x ) = x 2. The equation of the graph of y = x 2, can be written as y = f ( x ).

12. x This is the graph of y = f ( x ) + 1 and this is the graph of y = f ( x ) + 4. What do you notice? This is the graph of y = f ( x ) – 3 and this is the graph of y = f ( x ) – 7. What do you notice? Here is the graph of y = x 2, where y = f ( x ). y The graph of y = f ( x ) + b is the graph of y = f ( x ) translated by b units along the y-axis.

13. x This is the graph of y = f ( x – 1), and this is the graph of y = f ( x – 4). What do you notice? This is the graph of y = f ( x + 2), and this is the graph of y = f ( x + 3). What do you notice? The graph of y = f ( x - a ) is the graph of y = f ( x ) translated by a units along the x-axis. y Here is the graph of y = x 2 – 3, where y = f ( x ).

14. x This is the graph of y = – f ( x ). What do you notice? The graph of y = – f ( x ) is the graph of y = f ( x ) reflected in the x -axis. Here is the graph of y = x 2 –2 x – 2, where y = f ( x ). y

15. x Here is the graph of y = x 3 + 4 x 2 – 3 where y = f ( x ). y This is the graph of y = f (– x ). What do you notice? The graph of y = f (– x ) is the graph of y = f ( x ) reflected in the y -axis.

16. x This is the graph of y = 2 f ( x ). What do you notice? This graph is is produced by doubling the y - coordinate of every point on the original graph y = f ( x ). This has the effect of stretching the graph in the vertical direction. Here is the graph of y = x 2, where y = f ( x ). y The graph of y = pf ( x ) is the graph of y = f ( x ) stretched parallel to the y -axis by scale factor p.

17. x Here is the graph of y = x 2 + 3 x – 4, where y = f ( x ). y The graph of y = f ( ax ) is the graph of y = f ( x ) stretched parallel to the x -axis by scale factor a 1 This is the graph of y = f (2 x ). What do you notice? This graph is is produced by halving the x - coordinate of every point on the original graph y = f ( x ). This has the effect of compressing the graph in the horizontal direction.

18. We can now look at what happens when we combine any of these transformations. For example, since all quadratic curves have the same basic shape any quadratic curve can be obtained by performing a series of transformations on the curve y = x 2. Start by writing y = 2 x 2 + 4 x – 1 in completed square form: 2 x 2 + 4 x – 1 = 2( x 2 + 2 x ) – 1 = 2(( x + 1) 2 – 1) – 1 = 2( x + 1) 2 – 3 Write down the series of transformations that must be applied to the graph of y = x 2 to give the graph y = 2 x 2 + 4 x – 1.

19. So, starting with y = x 2 : y = x 2 y = ( x + 1) 2 y = 2( x + 1) 2 – 3 y = 2( x + 1) 2 1.Translate –1 units in the x -direction. 2.Stretch by a scale factor of 2 in the y -direction. 3.Translate –3 units in the y -direction. This is a quadratic of vertex form where the coordinates of the vertex is (-1,-3). We can now look at what happens when we combine any of these transformations.

21. y = ax 3 + bx 2 + cx + d (where a ≠ 0) A cubic function in x can be written in the form: Graphs of cubic functions have a characteristic shape depending on the values of the coefficients: When the coefficient of x 3 is positive the shape is When the coefficient of x 3 is negative the shape is or Cubic curves have rotational symmetry of order 2.

22. When a cubic function is written in the form y = a ( x – p )( x – q )( x – r ), it will cut the x -axis at the points ( p, 0), ( q, 0) and ( r, 0). p, q and r are the roots of the cubic function. When a cubic function is written in the form y = a ( x – p )( x – q )( x – r ), it will cut the x -axis at the points ( p, 0), ( q, 0) and ( r, 0). p, q and r are the roots of the cubic function. In general: To sketch the graph of a cubic function given in factorized form, Find the roots of the function and plot these on the x -axis. Find the y -intercept by putting x equal to 0 in the equation. Look at the coefficient of x 3 to decide whether the curve is N -shaped or -shaped.

23. y = x 3 – 4 x A cubic function always contains a term in x 3. It can also contain terms in x 2 or x or a constant. Here are examples of three cubic functions: y = x 3 + 2 x 2 y = -3 x 2 – x 3

24. A reciprocal function always contains a fraction with a term in x in the denominator. Here are examples of three simple reciprocal functions: In each of these examples the axes form asymptotes. The curve never touches these lines. y = 3 x 1 x –4 x

25. An exponential function is a function in the form y = a x, where a is a positive constant. Here are examples of three exponential functions: In each of these examples, the x -axis forms an asymptote. y = 2 x y = 3 x y = 0.25 x

26. Solve the equation 2 x 2 – 5 = 3 x using graphs. We can do this by considering the left-hand side and the right-hand side of the equation as two separate functions. 2 x 2 – 5 = 3 x y = 2 x 2 – 5 y = 3 x The points where these two functions intersect will give us the solutions to the equations.

27. –1–2–3–401234 –2 –4 –6 2 4 6 8 10 y = 2 x 2 – 5 y = 3 x (–1,–3) (2.5, 7.5) The graphs of y = 2 x 2 – 5 and y = 3 x intersect at the points: The x -value of these coordinates give us the solution to the equation 2 x 2 – 5 = 3 x as (–1, –3)and (2.5, 7.5). x = –1 and x = 2.5

28. Solve the equation 2 x 2 – 5 = 3 x using graphs. Alternatively, we can rearrange the equation so that all the terms are on the right-hand side, The line y = 0 is the x -axis. This means that the solutions to the equation 2 x 2 – 3 x – 5 = 0 can be found where the function y = 2 x 2 – 3 x – 5 intersects with the x -axis. 2 x 2 – 3 x – 5 = 0 y = 2 x 2 – 3 x – 5 y = 0

29. –1–2–3–401234 –2 –4 –6 2 4 6 8 10 y = 2 x 2 – 3 x – 5 y = 0 (–1,0) (2.5, 0) The graphs of y = 2 x 2 – 3 x – 5 and y = 0 intersect at the points: (–1, 0)and (2.5, 0). The x -value of these coordinates give us the same solutions x = –1 and x = 2.5

30. Solve the equation x 3 – 3 x = 1 using graphs. This equation does not have any exact solutions and so the graph can only be used to find approximate solutions. A cubic equation can have up to three solutions and so the graph can also tell us how many solutions there are. Again, we can consider the left-hand side and the right-hand side of the equation as two separate functions and find the x -coordinates of their points of intersection. x 3 – 3 x = 1 y = x 3 – 3 xy = 1

31. –1–2–3–401234 –2 –4 –6 2 4 6 8 10 y = x 3 – 3 x y = 1 The graphs of y = x 3 – 3 x and y = 1 intersect at three points: This means that the equation x 3 – 3 x = 1 has three solutions. Using the graph these solutions are approximately: x = –1.5 x = –0.3 x = 1.9

32. We can solve the equation |2 x – 5| = 3 by sketching the graph of y = |2 x – 5| and the graph of y = 3 and finding the points where they intersect. 0 y = 3 At point A, x = 1 y x At point B, x = 4 y = 2 x – 5 y = –(2 x – 5) A B

33. Solve the equations y = |2 x – 3| and y = | x – 4| by graphing both functions and noting the points where they intersect. 0 y = |2 x – 3| x y y = | x – 4| –12 A B So the solutions to |2 x – 3| = | x – 4| are x = –1 and x = 2.

34. We can also find the inverse of a one-to-one function f ( x ) by writing an equation of the form y = f ( x ) and rearranging it to make x the subject of the equation. For example, y ( x – 1) = 5 x xy – y = 5 x xy – 5 x = y x ( y – 5) = y Find the inverse of, x ≠ 1.

35. So the inverse function in terms of x is The inverse function is therefore of the form. Just replace the y ’s with x ’s. If we apply f ( x ) to a number and then apply f –1 ( x ) to the result we should get back to the original number. We can use this to check our result: as required. The domain f –1 ( x ) is x, x ≠ 5.