The document discusses key concepts in pre-calculus quadratics including: 1) Domains and ranges define the possible x- and y-values in quadratic functions. 2) The vertex identifies the peak or bottom of a parabola using the formula x = -b/2a, y = c. 3) The axis of symmetry is an imaginary line that divides the parabola exactly in half and is where x = the x-value of the vertex. 4) Completing the square converts quadratic functions into the form (x - h)2 + k to identify the vertex.

1. Pre-Calculus Quadratics By: Jennie Santos

2. Domains and Ranges Domain Values of “x” Usually goes on forever (- ∞ , ∞ ) Ranges Values of “y” Depends whether it is going up or down: If going up: (- #, ∞) If going down: (-∞, #)

3. Vertex To find the vertex use: f (x): a (x + b) ² + c Must have: “x” and “y” values X = opposite of “b” Y = “c” You can “transform” 0 = ax² + bx + c to find a vertex

4. Axis of Symmetry (A. of S.) …is an imaginary line that divides a parabola EXACTLY in half It is used to help find the vertex and other points in a parabola. A. of S. is what “x” is = to For example if the vertex is (-2, 5).. The A. of S. is x = -2

5. Completing a square How to convert y = ax ² + bx + c into y = a (x – h)² + k Eg. x² - 8 x + 16 Step 1: Change x² - 8x + 16 into (x² - 8x + ___) + 16 ^ In this blank spot you need to divide the middle term by 2 then square the number So it will be (x² - 8x + 16 ) + 16 -16 ^ Here you subtract the number outside of the bracket by 16 so it will keep it in balance. Step 2: (x² - 8x + 16 ) + 0 ^ Next you need to factor this into complete squares (x² - 8x + 16) + 0 = (x – 4)² + 0

6. Quadratic Problems Ex. 1) Making an equation Ex. 2) Finding an enclosed area Ex. 3) Given an equation

7. Making an equation A book store is opening & needs to know which price is best to sell a group of books, for the most profit. If they sell it at $10 each, 100 people will buy. For every $2 increase 5 less people will buy. What is the maximum revenue? Revenue = price · # sold Price -> 10+2x # sold -> 100-5x R = (10+2x)(100-5x) R = 1000 -50x+200x-10x² R = -10x²+150x+1000 R = -10(x²-15x+225/4)+4000/4+2250/4 R = -10(x-15/2) ²+6250/4 The maximum revenue is $1562.50. (You can check by using fitting : R = (10+2(15/2))(100-5(15/2)) and fitting x into where “x” is)

8. Ex. 2) Finding an enclosed area AREA = LENGTH x WIDTH = (620 – 2w) (w) = - 2w - 48050 * h = width / k = maximum area DIMENSIONS: LENGTH = 620 – 2w = 620 – 2(155) = 310m WIDTH = 155m EXAMPLE: At a local beach, the lifeguard has 620m of marker buoys to rope of a safe swimming area. Calculate the dimensions of the rectangular swimming area to create maximum swimming room if one side if the area is to be the beach. CHECK: K= maximum area Therefore : length x width = 48050m 310 x 155 = 48050 m

9. 3. Given an equation EXAMPLE: The height, h , in metres, after the launching of a rocket at any time, t , in seconds, is defined by the equation below. Find the maximum height reached by the rocket and the time it takes to reach this height. H = -3t + 9t + 81/4 = -3 (t -3t + 9/4) + 81/4 + 27/4 = -3 (t – 3/2) +27 *h = seconds / k= maximum height Therefore: Maximum height = 27m Seconds = 1.5