A Math Summary Booklet

QUADRATIC FUNCTIONS 01 Completing The Square Quadratic Inequalities Reverse Quadratic Inequality Application Step 1: Flush everything to the left and rearrange according to 𝑎𝑥2 + 𝑏𝑥 + 𝑐 Step 2: Simplify and rearrange according to 𝑎𝑥2 + 𝑏𝑥 + 𝑐 Step 3: Solve your quadratic inequalities Note: Always ensure 𝒙𝟐 is Positive, If it’s negative, divide and FLIP Your INEQUALITY SIGN. Step 1: Given that 𝑥 < −3 or 𝑥 > 2 Step 2: 𝑥 + 3 𝑥 − 2 > 0 (Reverse and Form Back Original) Step 3: 𝑥2 + 𝑥 − 6 > 0 (Expand) Maximum Point Starting Point Sub 𝑡 Sub 𝑦 = 0 −2𝑥2 + 4𝑥 + 8 = −2(𝑥2 − 2𝑥 − 4) = −2[(𝑥2 − 2𝑥 + (−2 2 )2 − (−2 2 )2 − 4] = −2[(𝑥 − 1)2 − 5] = −2(𝑥 − 1)2 + 10 When 𝒙𝟐 𝒊𝒔 𝒏𝒐𝒕 𝟏 Reverse Inequalities

Simultaneous Equations CHAPTER 1: QUADRATIC FUNCTIONS 02 Algebra Word Problems Validation Important Concepts𝑥2 + 𝑦2 = 34 … (1) 𝑦 + 3𝑥 = 14 … (2) Using (2) 𝑦 = 14 − 3𝑥 Substitute into (1) 𝑥2 + 14 − 3𝑥 2 = 34 𝑥2 + 196 − 84𝑥 + 9𝑥2 = 34 10𝑥2 − 84𝑥 + 162 = 0 𝑥 − 3 5𝑥 − 27 = 0 𝑥 = 3 or 𝑥 = 27 5 Substitute into (2) 𝑦 + 3 3 = 14 𝑦 + 3 27 4 = 14 𝑦 = 14 − 9 = 5 𝑦 = 14 − 3 27 4 = − 11 5 Answer: 𝑥 = 3, 𝑦 = 5 𝑥 = 5 2 5 , 𝑦 = −2 1 5 The line 2𝑥 + 3𝑦 = 8 meets the curve 2𝑥2 + 3𝑦2 = 110 at the point A and B. Find the coordinates of 𝐴 and B. 2𝑥 = 8 − 3𝑦 𝑥 = 8 − 3𝑦 2 Substitute into (2) 2 8 − 3𝑦 2 2 + 3𝑦2 = 110 2 64 − 48𝑦 + 9𝑦2 4 + 3𝑦2 = 110 128 − 96𝑦 + 18𝑦2 + 12𝑦2 = 440 30𝑦2 − 96𝑦 − 312 = 0 10𝑦2 − 32 − 104 = 0 𝑦 + 2 5𝑦 − 26 = 0 𝑦 = −2, 𝑦 = 26 5 Substitute into (1) 𝑥 = 8−3(−2) 2 = 7 𝑥 = 8−3(26 5 ) 2 = − 19 5 Answer: 7, −2 and (−3 4 5 , 5 1 5) For Simultaneous Equations, • Validate by Substituting Your Final Answer back into the Original Question • If the question is related to coordinates, ensure that you leave your answers in (𝑥, 𝑦) Concept: • There are 2 methods to solve for Simultaneous, either Substitution Method or Elimination Method. • I highly recommend to use Substitution Method as I find that it is faster and easier.

Completing The Square CHAPTER 1: QUADRATIC FUNCTIONS 03 Easy Advance Validation Important ConceptsSimplify 𝑥2 + 4𝑥 − 12 = 𝑥2 + 4𝑥 + 4 2 2 − 4 2 2 − 12 = 𝑥 + 2 2 − 16 Hence, Solve 𝑥2 + 4𝑥 − 12 = 0 𝑥 + 2 2 − 16 = 0 𝑥 + 2 2 = 16 𝑥 + 2 = 4 or 𝑥 + 2 = −4 • After you get your final answer, re-expand back to make sure it gives you back the original answer. • If you are solving, you can check your solution by using the Quadratic Equation Function in your calculator... they should be the same. Concept: 1. The condition for Completing The Square is that the coefficient of 𝑥2 MUST BE +1. If it is not +1, we need to FACTORISE the value to make it +1. 2. Be careful of the values you substitute in the bracket. Always include the SIGN. 3. When solving and completing the square, always solve by Square Rooting the values. NEVER expand back and solve by factorisation. That defeats the purpose of Completing The Square. Simplify 𝑥2 − 6𝑥 + 8 = 𝑥2 −6𝑥 + − 6 2 2 − − 6 2 2 + 8 = 𝑥 − 3 2 − 1 Hence, Solve 𝑥 − 3 2 − 1 = 0 𝑥 − 3 2 = 1 𝑥 − 3 = 1 or 𝑥 − 3 = −1 𝑥 = 4 or 𝑥 = 2 Simplify −𝑥2 − 𝑥 + 2 = −(𝑥2 + 𝑥 − 2) = −[(𝑥2 + 𝑥 + 1 2 − 1 2 − 2] = −[ 𝑥 + 1 2 2 − 5 2] = − 𝑥 − 1 2 2 + 5 2 Simplify 2𝑥2 − 5𝑥 + 9 = 2(𝑥2 − 5 2 𝑥 + 9 2) = 2[(𝑥2 − 5 2 𝑥 + − 5 4 2 − − 5 4 2 + 9 2] = 2[ 𝑥 − 5 4 2 + 47 16] = 2 𝑥 − 5 4 2 + 47 8

Method 2: Completing the Square Graphical Methods CHAPTER 1: QUADRATIC FUNCTIONS 04 Method 1: Fully Factorised Validation Important ConceptsSketch 𝑦 = 𝑥2 + 3𝑥 + 2 1) Find the Roots 𝑆𝑢𝑏 𝑦 = 0 𝑥2 + 3𝑥 + 2 = 0 𝑥 + 2 𝑥 + 1 = 0 𝑥 = −1 𝑜𝑟 𝑥 = −2 2) Find the y-intercept 𝑆𝑢𝑏 𝑥 = 0 𝑦 = 2 3) Find the turning point 𝑆𝑢𝑚 𝑜𝑓 𝑅𝑜𝑜𝑡𝑠 2 = −1 + −2 2 = −1.5 𝐿𝑖𝑛𝑒 𝑜𝑓 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑦 𝑥 = −1.5 𝑆𝑢𝑏 𝑥 = −1.5 𝑦 = −1.5 2 + 3 −1.5 + 2 = −0.25 Turning point (−1.5, −0.25) Sketch 𝑦 = 𝑥 + 2 2 − 9 1) Find the turning point (−2, −9) 2) Find the y-intercept 𝑆𝑢𝑏 𝑥 = 0 𝑦 = 2 2 − 9 = −5 3) Find the roots 𝑆𝑢𝑏 𝑥 = 0 𝑥 + 2 2 − 9 = 0 𝑥 + 2 2 = 9 𝑥 + 2 = 3 𝑜𝑟 𝑥 + 2 = −3 𝑥 = 1 𝑜𝑟 𝑥 = −5 Concept: There are two types of graph sketching. 1) Fully Factorised Equation 2) Completing The Square Equation The steps are different due to the ease & convenience of finding the points. When drawing a quadratic graph, I’m interested in knowing 3 things: 1) Roots 2) Y Intercept 3) Turning Point Finding Roots and Y Intercepts are similar for both types of graphs. However, the key difference is in finding the Turning Point. Look at how I obtained the turning point for both methods! Careless: Coefficient of 𝑥2 - Happy or Sad Face Coordinates (𝑥, 𝑦) vs Value – Number Line of Symmetry - Equation After you sketch your graph, make sure that the values make sense. Curve, Turning Point, Roots, Y Intercept,

Applications CHAPTER 1: QUADRATIC FUNCTIONS 05 Word Problems Validation Important ConceptsThe path of a water jet can be modelled by the quadratic function 𝑦 = 𝐶 𝑥 − 1.2 2 + 2.25, where x m is the horizontal distance it travels, y m is the height of the water above the ground and C is a constant. The initial height of the water jet is 1.05 m above the ground. (i) Find the value of C. (ii) Find the maximum height above the ground that the water jet reaches. (iii) Find the value of x for which the water jet is 1.05m above the ground again. (iv) Find the maximum horizontal distance travelled by the water jet (i) 𝑦 = 𝐶 𝑥 − 1.2 2 + 2.25 Sub 𝑥 = 0, 𝑦 = 1.05 1.05 = 𝐶 −1.2 2 + 2.25 −1.2 = 𝐶 1.44 𝐶 = − 5 6 or −0.833 (ii) 2.25mm (iii) 𝑦 = − 5 6 𝑥 − 1.2 2 + 2.25 Sub 𝑦 = 1.05 1.05 = − 5 6 𝑥 − 1.2 2 + 2.25 −1.2 = − 5 6 𝑥 − 1.2 2 1.44 = 𝑥 − 1.2 2 1.2 or −1.2 = 𝑥 − 1.2 𝑥 = 2.4 or 0(NA) (iii) 𝑦 = − 5 6 𝑥 − 1.2 2 + 2.25 Sub 𝑦 = 0 0 = − 5 6 𝑥 − 1.2 2 + 2.25 −2.25 = − 5 6 𝑥 − 1.2 2 2.7 = 𝑥 − 1.2 2 2.7 or − 2.7 = 𝑥 − 1.2 𝑥 = 2.84 or −0.443(𝑁𝐴) Max horizontal distance = 2.84m A support cable for a bridge is parabolic in shape. The cable is supported by 25 m tall towers A and B that are 80 m apart. Vertical supporting wires are spread out in equal intervals hanging from cable. The lowest point on the cable is 5 m above the roadbed. The height of the cable above the roadbed is given as y m and the horizontal distance from Tower A is given as x m. (i) Find a quadratic function in the form 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 to model this situation. (ii) Find the length of the vertical supporting wire that is 15 m horizontally from the origin. At lowest point, (40,5) So, 𝑦 = 𝑎 𝑥 − 40 2 + 5 When 𝑥 = 0, y= 25 25 = 𝑎 0 − 40 2 + 5 𝑎 = 1 80 When 𝑥 = 15, 𝑦 = 1 80 15 − 40 2 + 5 𝑦 = 12.8125 Length of the wire is 12.8125 m Accept 12 13 16 m Word Problems Concept: Many students struggle with this because it feels odd and challenging. However, this portion is just applying the concepts from Completing The Square. Under Completing The Square, we learn a few things: 1) You can only complete the square if the coefficient of 𝑥2 is +1. 2) Obtaining Turning Points (Line of Symmetry, Maximum and Minimum Value) 3) Solving Completing The Square via Square root Method and not Quadratic Formula Sit down 15 minutes, internalise this and you will definitely get it right! Validation of Completing The Square requires you to expand back to double check if it gives you the original equation. I typically will do this before continuing with the question because I don’t want to risk redoing the whole question if I make a mistake in my completing the square steps.

Quadratic Inequalities Quadratic Inequalities CHAPTER 2: EQUATION AND INEQUALITIES 06 Reverse Quadratic Inequalities Validation Important ConceptsSolve 𝑥2 + 3𝑥 + 2 > 0 𝑥2 + 3𝑥 + 2 > 0 𝑥 + 2 𝑥 + 1 > 0 𝑥 < −2 or 𝑥 > −1 Solve 𝑥2 + 3𝑥 + 2 < 0 𝑥2 + 3𝑥 + 2 < 0 𝑥 + 2 𝑥 + 1 < 0 −2 < 𝑥 < −1 Find the value of 𝑏 for which −2 < 𝑥 < 1 3 is the solution of 3𝑥2 + 5𝑥 < 𝑏. −2 < 𝑥 < 1 3 𝑥 + 2 3𝑥 − 1 < 0 3𝑥2 − 𝑥 + 6𝑥 − 2 < 0 3𝑥2 + 5𝑥 − 2 < 0 3𝑥2 + 5𝑥 < 2 𝑏 = 2 Find the value of 𝑏 for which 𝑥 < −2 𝑜𝑟 𝑥 > 1 3 is the solution of 3𝑥2 + 5𝑥 > 𝑏. 𝑥 < −2 𝑜𝑟 𝑥 > 1 3 𝑥 + 2 3𝑥 − 1 > 0 3𝑥2 − 𝑥 + 6𝑥 − 2 > 0 3𝑥2 + 5𝑥 − 2 > 0 3𝑥2 + 5𝑥 > 2 𝑏 = 2 Solve −𝑥2 + 3𝑥 − 2 > 0 ** −𝑥2 + 3𝑥 − 2 > 0 𝑥2 − 3𝑥 + 2 < 0 𝑥 − 2 𝑥 − 1 < 0 1