Quadratic Functions Lesson #5

Transcription

1 Quadratic Functions Lesson #5 Axes Of Symmetry And Vertex Axis Of Symmetry As we have seen from our previous exercises: o the equation of the axis of symmetry of y = ax + bx+ c is x =. a The problem with the method used to demonstrate this is that not all quadratic functions have a graph which cuts the x-axis. To prove this formula we can use an expansion method and compare coefficients. Proof: Suppose y = ax + bx+ c is converted to y = a(x h) + k. i.e., y = a(x hx+ h ) + k i.e., y = ax ahx + [ah + k] Comparing the coefficients of x we obtain: ah = b. Therefore, h =. a Example: Find the equation of the axis of symmetry of y = x + 3x+ 1. y = x + 3x+1 has a =, b = 3, c = 1 Therefore, axis of symmetry has equation i.e., 3 x =. 4 Turning Point (Or Vertex) x 3 = =. a The turning point (or vertex) of any parabola is the point at which the function has a maximum value (for a< 0) or, a minimum value (for a > 0).

2 As the turning point lies on the axis of symmetry, its x-coordinate will be x =. a The y-coordinate can be found by substituting for x in the function, i.e., f a. Example 1: Determine the coordinates of the vertex of y = x 8x+ 1. y = x 8x+1 has a =, b = 8, and c = 1, ( 8) so = =. a Therefore, the axis of symmetry is x =. When x =, y = () 8() + 1 y = y = 7 Therefore, the vertex has coordinates (, -7). Example : For the quadratic function y = x + x+ 3: (a.) find its axes intercepts (b.) find the equation of the axis of symmetry (c.) find the coordinates of the vertex (d.) sketch the function showing all important features. (a.) When x = 0, y = 3. Therefore, the y-intercept is 3. When y = 0, x + x+ 3 = 0 x x 3 = 0 (x 3)(x + 1) = 0 So, x = 3 or x = 1. Therefore, the x-intercepts are 3 and -1. (b.) a = 1, b =, and c = 3 So, = = 1. a Therefore, the axis of symmetry is x = 1.

3 (c.) From (b.), when x = 1, y = (1) + (1) + 3 y = y = 4 Therefore, the vertex is (1, 4). (d.) Where Functions Meet Consider the graphs of a quadratic function and a linear function on the same set of axes. Notice that we could have: ( points of intersection) (1 point of intersection) (no points of intersection) The graphs could meet and the coordinates of the points of intersection of the graphs of the two functions can be found by solving the two equations simultaneously. Example: Find the coordinates of the points of intersection of the graphs with equations y = x x 18 and y = x 3. y = x x 18 meets y = x 3 where x x 18 = x 3 x x 15 = 0 {RHS = 0 } (x 5)(x + 3) = 0 {factorising} So, x = 5 or x = 3. Substituting into y = x 3, when x = 5, y = and when x = 3, y = 6.

4 Therefore, the graphs meet at (5, ) and (-3, -6). Quadratic Modelling There are many situations in the real world where the relationship between two variables is a quadratic function. This means that the graph of such a relationship will be either or and the function will have a minimum or maximum value. o For y = ax + bx+ c: if a > 0, the minimum value of y occurs at x = a if a < 0, the maximum value of y occurs at x =. a The process of finding the maximum or minimum value of a function is called optimisation. Optimisation is a very useful tool when looking at such issues as: o maximising profits o minimising costs o maximising heights reached etc. Example 1: The height H metres, of a rocket t seconds after it is fired vertically upwards is given by H(t) = 80t 5t, t > 0. (a.) How long does it take for the rocket to reach its maximum height? (b.) What is the maximum height reached by the rocket? (c.) How long does it take for the rocket to fall back to earth? (a.) H(t) = 80t 5t H(t) = 5t + 80t, where a = 5, b = 80 and c = 0.

5 Since a = 5, the shape is. The maximum height reached occurs when t =, a 80 i.e., t = ( 5). So, t = 8. Therefore, the maximum height is reached after 8 seconds. (b.) H(8) = (8) H(8) = H(8) = 30 i.e., the maximum height reached is 30 m. (c.) The rocket falls back to earth when H(t) = 0, 0 = 80t 5t 5t + 80t = 0 5t(t 16) = 0 {factorising} So, t = 0 or t = 16. i.e., the rocket falls back to earth after 16 seconds. Example : A vegetable gardener has 40 m of fencing to enclose a rectangular garden plot where one side is an existing brick wall. If the two equal sides are x m long: (a.) show that the area enclosed is given by A = x(40 x) m (b.) find the dimensions of the vegetable garden of maximum area. (a.) Side XY = 40 x m. Now, area = length width, so, A = x(40 x) m. (b.) A = 40x x = x + 40x is a quadratic in x, with a =, b = 40 and c = 0. Since a< 0, the shape is. So, maximum area occurs when 40 x = = = 10. a 4

6 Therefore, area is maximized when YZ = 10 m and XY = 0 m. Example 3: A manufacturer of pot-belly stoves has the following situation to consider. 400 If x are made per week, each one will cost 50 + x dollars and the total receipts per week for selling them would be 550x x dollars. ( ) How many pot-belly stoves should be made per week in order to maximise profits? Total profit, P = receipts costs 400 So, P = ( 550x x ) 50 + x x P = 550x x 50x 400 P = x + 500x 400 dollars which is a quadratic in x, with a =, b = 500 and c = 400. Since a< 0, the shape is. 500 So, P is maximised when x = = = 15 a 4 Therefore, produce 15 of them per week.