MPM 2D0 Unit 4: Graphing Quadratics REVIEW Graphing Quadratics Base function is y = x2. From (0, 0) go: over 1 up 1, over 2 up 4, over 3 up 9, over 4 up 16... Terms: Quadratic Function, Parabola, Axis of Symmetry, Vertex, Maximum/Minimum Translations of y = x2 y a x h k 2 a = reflection if negative, vertical stretch if a > 1, vertical compression if 0 < a < 1 h = horizontal translation left if h > 0 (positive), right if h < 0 (negative) k = vertical translation up if k > 0 (positive), down if k < 0 (negative) Draw base graph, do stretch/compression and reflection, move each point to new vertex OR start at new vertex and build graph Example 1: State the transformations for each of the following, then graph using transformations. Complete the chart provided. b) y a) y x 4 2 Down 4 1 x 22 2 c) y 2 x 6 8 2 Vertical Compression by a factor of ½, right 2 y = x2 Reflection in the x-axis, Vertical Stretch by a factor of 2, left 6, up 8 y = x2 y = –2(x + 6)2 + 8 y = x2 – 4 y = ½ x2 y = x2 y = ½ (x – 2)2 y = –2x2 Property Vertex Axis of Symmetry Stretch or compression factor relative to y x 2 Direction of opening Values x may take Values y may take a (0, –4) x=0 b (2, 0) x=2 C (–6, 8) x = –6 None ½ 2 (reflection) Up Set of real numbers y ≥ –4 Up Set of real numbers y≥0 Down Set of real numbers y≤8 Graphing Quadratics With Roots y a( x r )( x s) The x-intercepts (or zeros, roots) are r and s. MPM 2D0 Find x-intercepts by setting y = 0, then solving using factoring or quadratic formula. Find y-intercept by setting x = 0 and solving for y (or is constant term in standard form). Find x-value of vertex by adding the two x-intercepts and dividing by two (half-way point). Find y-value of vertex by substituting x-value of vertex into original equation. Plot points and join with a smooth curve. Example 2: Sketch y 2 x 4 x 6 using the x-intercepts, y-intercept, and vertex. 2 x-intercepts Vertex Set y = 0 x-value 0 2 x 4 x 6 3 1 2 2 2 1 2 0 2 x 2 2 x 3 0 2 x 3 x 1 x 3 0 x 1 0 x 3 x 1 y-value y-intercept y 2 x 2 4 x 6 y6 y 2(1) 2 4(1) 6 y 2 4 6 y 8 Determine the Equation From a Graph Forms of quadratic functions: o Vertex Form: y a x h k o Intercept Form: y a( x r )( x s) 2 o Standard Form: y ax bx c 2 Identify the vertex from the graph. Substitute into y a x h k . 2 Count to identify the a-value, OR pick another point and substitute into y a x h k and solve for "a" Convert equation to standard form if required. 2 Example 3: Determine an equation for the given parabola, in vertex form. 3 9a y a x h k 2 y ax 3 4 2 Use (0, 1): 1 a0 3 4 1 4 9a 2 3 a 9 1 a 3 y 1 x 32 4 3 MPM 2D0 Completing the Square Used to convert equations in standard form to vertex form Group x-terms. Factor out a-value from x-terms (not constant term) Divide the coefficient of the middle term by 2, square it, then add and subtract that number inside the brackets. Remove the subtracted term from the brackets. On the way out of the brackets, it is multiplied by the a-value. Write the perfect square trinomial as the square of a binomial. Example 4: Convert each of the following to vertex form. a) y x 2 6 x 3 b) y x 2 8 x 11 y x y x y x2 6x 3 6x 9 9 3 2 6x 9 9 3 y x 3 6 2 2 y x 2 8 x 11 y 3 x 2 12 x 9 y x 2 8 x 11 y 3 x 2 4 x 9 y x y x 2 c) y 3 x 12 x 9 2 8 x 16 16 11 2 8 x 16 16 11 y 3x y 3x 2 4x 4 4 9 2 4 x 4 12 9 y 3 x 2 3 y x 4 5 2 2 Word Problems Look for words like "maximum/minimum." You need the vertex, so you will have to complete the square if the equation is in standard form. You will also need to find points like the y-intercept and the x-intercepts. Example 5: The height, h metres, of a batted baseball as a function of the time, t seconds, since the ball was hit can be modeled by the function h 2.1t 10.08t 0.904 a) What was the maximum height of the ball? b) What was its height when it was hit, to the nearest tenth of a metre? c) How many seconds after it was hit did the ball hit the ground, to the nearest tenth of a second? d) What was the height of the ball, to the nearest tenth of a meter, 1s after it was hit? 2 a) h 2.1t 2 10.08t 0.904 b) y-int = 0.904 The height when it was hit was 0.9 m. 2.1t 2 10.08t 0.904 2.1t 2.1t 2.1 t 2 4.8t 0.904 c) Set h = 0 and solve for t. 2 4.8t 5.76 5.76 0.904 2 4.8t 5.76 12.096 0.904 2.1t 2.4 13 2 Vertex: (2.4, 13) t h Therefore, the maximum height was 13 m. 0 2.1t 2 10.08t 0.904 10.08 10.082 4(2.1)(0.904 ) t 2(2.1) 10.08 109 .2 4 .2 t 0.1 t 4 .9 inadmissib le Therefore, the ball was in the air for 4.9 s. MPM 2D0 d) Set t = 1, and solve for h h 2.1t 2 10.08t 0.904 2.1(1) 2 10.08(1) 0.904 2.1 10.08 0.904 Therefore, the ball was 8.9 m high 1 second after it was hit. 8.884 Finite Differences If the first differences are constant, the function is linear If the second differences are constant, the function is quadratic If neither the first nor second differences are constant, the function is neither linear nor quadratic Example 6: Use finite differences to determine whether the function is linear, quadratic, or neither. x 0 1 2 3 4 y –5 –6 –1 10 27 1st Diff 2nd Diff –6 – (–5) = –1 –1 – (–6) = 5 10 – (–1) = 11 27 – 10 = 17 5 – (–1) = 6 11 – 5 = 6 17 – 11 = 6 Since the 2nd differences are the same, the function is quadratic.