Quadratics NOTES.notebook November 02, 2017
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1 1) Find y where y = 2-1 and a) = 2 b) = -1 c) = 0 2) Epand the brackets and simplify: (m + 4)(2m - 3) To find the equation of quadratic graphs using substitution of a point. 3) Fully factorise 4y 2-5y - 6 Quadratic Graphs A quadratic equation involves a squared term e.g = 0 The graph of y = k 2 The simplest quadratic graph is y = 2 happy sad Positive k Negative k y = 2 graph is stretched by a factor of k e.g. Find the equation of the graph of the form y = k 2 1) Factorise fully: ) Without a calculator, find ) Without a calculator, simplify
2 1) Fully factorise: 3m m + 9 1) Multiply out the brackets: (e - 1)(2e + 1) 2) Simplify the following: 2) Factorise the epression 2gh - 8h 2 3) Without a calculator, find ) Evaluate How does the graph of y = 2 compare with the graph of y = 2 + 1? What about y = 2 + 2? y = 2 + q What about y = 2-2? positive q negative q e.g. Find k and q from the graphs of y = k 2 + q: 1) y y (-2, 15) (0,3) (0,-1) (2,-13) 2) 1 2 1) Write m in surd form. 2) Multiply out the brackets: (3e - 2)(e + 4) 3) Factorise the epression n 2-3n - 10
3 The graph of y = ( + p) 2 e.g. Find p for these graphs of y = ( + p) 2 : a) y b) y positive p negative p (6,0) (-5,0) 1) Epand the brackets: (2k - 3)(k + 3) 2) Factorise the epression 3n 2 + 2n - 1 3) Write down all the factors of 100 To sketch graphs of the form y = ( + a) 2 + b 4) Write down the gradient and y-intercept of the straight line 3y + 12 = 2 Sketching Quadratic Graphs We can be asked to label: Turning Point and its nature Roots (where it crosses the -ais) e.g. 2) Sketch the graph of y = ( + 1) 2-5 y-intercept Equation of the ais of symmetry e.g. 1) Sketch y = -( + 3) 2 and label all of the above.
4 Sketching quadratic graphs. To sketch graphs of the form y = ( + a)( + b) If a b = 0, then what can we say about a and b? ( + 4)( - 2) = 0 Sketching Quadratic Graphs These equations aren't in completed square form. e.g. 1) Sketch the graph y = ( - 1)( - 5) 2) Sketch the graph y = -( + 2)( - 2)
5 1. Factorise: Factorise: Factorise: State the gradient of the line: 4y + 12 = 2 Fully factorise the following: 1) m 2 + 2m - 8 2) 3p 2-13p ) 2p ) 4m 2-4m - 15 Sketch graphs, labelling any roots, the y-intercept, the equation of the line of symmetry, as well as the coordinates of the TP and its nature. 1) 6) 2) 7) 3) 8) 4) 9) 5) 10) Solve the equation ( + 4)( - 1) = 0 for. To solve quadratic equations. How might we solve = 0?
6 Match equivalent epressions... (m + 3) 4 (m - 2) (m + 3) 3 (m - 2) 2 (m + 4)(m - 4) Solving Quadratic Equations m 2-16 A quadratic equation can be written as a 2 + b + c = 0 Then, we can solve by factorising. c 0 1 Eamples: m 2 + 5m + 6 m 2-4 1) = 0 2) = 0 c 1 2 c Solve the quadratic equations: 1. a) ( + 5) = 0 b) ( - 3)( + 5) = 0 c) 2( - 10) = 0 d) ( - 6)( + 6) = 0 Solve the quadratic equations: 1. a) ( + 5) = 0 b) ( - 3)( + 5) = 0 c) 2( - 10) = 0 d) ( - 6)( + 6) = 0 2) a) = 0 b) 2-9 = 0 c) = 0 d) = 0 e) = 0 f) = 0 g) = 0 h) = 0 i) = 0 2) a) = 0 b) 2-9 = 0 c) = 0 Eample: Solve = 0 d) = 0 e) = 0 f) = 0 g) = 0 h) = 0 i) = 0
7 Solve these quadratic equations: 1) = 0 2) = 0 Solve these quadratic equations: 1) = 0 2) = 0 3) = 0 4) = 0 5) = 0 3) = 0 4) = 0 5) = 0 How can we solve these equations? 1) = 0 2) = 0 3) = 0 To write any quadratic equation in the form a 2 + b + c = 0 and to solve equations that don't factorise by using the quadratic formula. 4) 2 = 5 5) = -10 The Quadratic Formula If we have an equation a 2 + b + c = 0 that we can't factorise, we can use the Quadratic Formula to find solutions: Solve the quadratic equations by factorising: 1) = 0 2) = 0 3) 4-2 = 0 Eamples: 1) (given in eams) 4) = 5 2)
8 Solve the equations to 1 decimal place: Solve the equations to 1 decimal place: How can we tell how many roots an equation has? The Discriminant For a quadratic equation a 2 + b + c = 0 the discriminant is b 2-4ac. b 2-4ac > 0 means 2 real, distinct roots b 2-4ac = 0 means 2 real, equal roots b 2-4ac < 0 means no real roots e.g. 1) Determine the nature of the roots of 2( + 1) = 2-3 e.g. 2) Find the range of values for T such that T = 0 has 2 real, distinct roots. y = ( - 1) 2-25 Spot the mistake(s)! y Roots: 0 = = ( - 4)( + 6) = 4 or -6 (-6,0) (4,0) y intercept: y = (-1) 2-25 = -26 Equation of ais of symmetry: = -1 (0,-26) (-1,-25) TP occurs at (-1, -25) and is a minimum because 2 >0
9 The areas of these rectangles are equal. a) Find the value of. b) Calculate the area of the rectangles. To solve problems using quadratics. (2 + 2) cm ( + 1) cm ( + 3) cm ( + 4) cm Find a for the right angled triangle.
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