1. Quadratic Formula Sam Scholten

2. Graphing Standard Form Graphing Standard form: Standard form in Quadratic functions is written as: Y = ax 2 +bx+c. The parabola will open up if a>0 and down if a 1. The x coordinate of the vertex is –b/2a. The axis of symmetry is the vertical line x = -b/2a. Y = x 2 :

3. Examples Y = 2x 2 – 8x + 6 The coefficients for this function are a = 2, b = -8 and c = 6. Since a > 0, the parabola opens up. To find the vertex, the x coordinate is: x = 8/4 or 2. The y coordinate is: 2(2) 2 – 8(2) + 6 or -2. The vertex is (2,-2) and the axis of symmetry is x = 2. Go up 2 and over 1 on each side of the symmetry so you get the points (1,0) and (3,0). Go up 6 more and 1 more over on both sides to get (0,6) and (4,6) (since is 2 up and over 1 the first time, you get 3 times the y increase the second time, but no change on the x). Draw you parabola through the points.

4. RESULT

5. Graphing Vertex Form Graphing Vertex Form: In vertex form, a quadratic function is written as: y = a(x-h) 2 + k, where the vertex is (h, k) and the axis of symmetry is x = h. Y = -1/2(x + 3) 2 + 4:

6. EXAMPLES Y = -1/2(x + 3) 2 + 4 a = -1/2, h = -3 and k = 4 and since a < 0, the parabola opens down. To graph, plot the vertex (h,k) which is (-3,4) and draw the axis of symmetry which is x = -3. Since it is -1/ 2, we will be going negative two down and two to the side on both sides. Then you multiply the y axis by three, so six more down and two more over.

7. Result

8. Graphing Intercept Form Graphing Intercept Form: The intercept form of a quadratic function is y= a(x-p)(x-q), in this formula, the P and the Q are x-intercepts and the axis of symmetry is halfway between (P,0) and (0,Q). Y = -(x + 2)(x – 4):

9. EXAMPLES Y = -(x + 2)(x – 4) a = -1, p = -2 and q = 4 The x intercepts and p and q so you go halfway between them to find the axis of symmetry which is x = 1. The x coordinate of the vertex is one, so we plug that into the equation and get nine for the y coordinate. Our vertex is (1,9). You go down one and over one for the first point on both sides. Then multiply the y axis by three and you go down three more and over one more. ( you can also think of it as adding to the y axis the number two times the first y coordinate. EX: (1,4) being the first, the next point would be 2,12 and the next would be (3,20)

10. RESULT

11. Solving Quadratics With Square Roots Square roots: The product property of square roots is written as: √ab = √a * √b The quotient Property is: √a/b = √a/√b provided that in both, a>0 and b>0. The square root of a negative number (if r is a real positive number) is shown as i√r which means (i√r) 2 = -r Square roots can be used in many ways in math. If s > 0, then x2 = s has two solutions. x = √s and x = -√s which will often be written like x = +√s which means both. (i stands for imaginary unit)

12. EXAMPLE1 3x 2 + 10 = -26 Subtract 10 from both sides (3x 2 = -36) Divide by 3 (x 2 = -12) Take the squares of both sides (x = +√-12) Write in terms of i (x = +i√12) Simplify the radical (x= +2i√3) The solutions are -2i√3 and 2i√3

13. EXAMPLE 2 -3x 2 + 6 = 15 Subtract 6 from both sides ( -3x 2 = 9) Divide by -3, (x 2 = -3) Take the squares (x = √-3) Terms of i, (x = +i√-3) The solutions are -i√-3 and i√-3

14. More Complex Numbers Standard form of a complex number is written as: a + bi where a is the real number and bi is the imaginary part. In a complex plane, the horizontal axis is all the real numbers, and the vertical axis is complex numbers. These are called the Real axis, and the Imaginary axis. Plot: 2 – 3i, -3 + 2i, and 4i: (-3 + 2i) (0,4i) (2 – 3i)

15. Even More Complex Numbers Complex numbers that are written as: a + bi then a – bi, are called complex conjugates, which, when multiplied together, form a real number. You can also use the conjugates to show the quotient of two complex numbers in standard form. The absolute value of a complex number is shown as: z = a + bi, and |z| is a non-negative real number defined as: |z| = √(a 2 + b 2 ).

16. Solving Quadratics By Factoring Solving quadratics with factoring: a 2 - b 2 = (a + b)(a – b) is the difference of two squares. A perfect square trinomial is written as: a 2 + 2ab + b 2 = (a + b) 2 or: a 2 - 2ab + b 2 = (a - b) 2 The zero product property is: if A and B are real numbers or algebraic expressions and AB = 0 then either A or B will be equal to 0. In a standard quadratic equation, x 2 + bx + c = (x + m)(x + n) = x + (m + n)x + mn The sum of n and m must be equal to b and the product must be equal to c

17. Example 1 3x 2 – 17x + 10 You want 3x 2 – 17x + 10 to equal (kx + m) (lx + n) where l and k are factors of 3 and m and n are (negative) factors of 10. There are several options but only (3x – 2)(x – 5) is correct.

18. Special patterns 1. The difference of 2 squares: 4x 2 – 25 = (2x) 2 – 5 2 equals (2x + 5)(2x – 5) 2. Perfect square trinomial: 9y 2 + 24y + 16 = (3y) 2 + 2(3y)(4) + 4 2 equals (3y + 4) 2

19. Solving Quadratics By Completing The Square Completing the square is a process that allows you to write the form x 2 + bx. This models the following rule: x 2 + bx + (b/2) 2 = (x + b/2) 2 x x bx x (b/2)x x2x2

20. Examples 1. x 2 – 7x + c You get c by taking half of b which in this case is -7 c = (b/2) 2 = ( -7/2) 2 = 49/4 So the original equation is now x 2 – 7x + 49/4 So the square is now (x - 7/2) 2 2. x 2 + 10x – 3 = 0, add 3 to both sides x 2 + 10x = 3, add (10/2) 2 to both sides x 2 + 10x + 5 2 = 3 + 25, write the left as a binomial squared (x + 5 ) 2 = 28, take the square roots of both sides x + 5 = +√28, solve for x x = -5 + 2√7, the solutions are x = -5 + 2√7 and x = -5 – 2√7

21. Solving By Using The Quadratic Formulas There are 4 types of quadratic inequalities Y ax 2 + bx + c Y > ax 2 + bx + c Y < ax 2 + bx + c Draw the parabola making sure to have a dashed line for and and < Choose a point inside the parabola and check whether or not it’s a solution. If the point is a solution then shade the region inside the Parabola, if not, shade the outside.

22. Example 1&2 y > x 2 – 2x – 3 Since its >, the parabola is dashed instead of solid. Test the point (1,0) in the equation. 0 > 1 – 2 – 3 is 0 > -4 Which is a solution. The graph will be a dotted line with it shaded inside with a vertex of (1, -4) y < -2x 2 – 2x – 3 Since its <, the parabola is solid. Test the point (1, -6) which would make -6 < -2 – 2 – 3 is -6 < -7 which is not a solution making the graph be a solid line opening down and wider than a standard graph with a vertex of (1, -4) and shaded on the outside.

23. EXAMPLE 3 y > 2x 2 – 2x – 3 Since it is > this will be a dashed line. Plug in (2,4) which makes: 4 > 8 – 4 – 3 is 4 > 1 which is true which makes it shaded on the inside of a dashed graph with a vertex of (1, -4)

24. Word Problem You want to plant a rectangular garden along part of a 40 foot side of your house. To keep out animals, you will enclose the garden with wire mesh along its three open sides. You will also cover the garden with mulch. You have 50 ft of mesh and enough mulch to cover 100 sq. ft. What should the garden’s dimensions be?

25. Solution X will be the sides perpendicular to the house and 50 – 2x is the length of the third side. X(50 – 2x) = 100,Length * width = area 50x – 2x 2 = 100,distributive property -2x 2 + 50x = 100,Write the x 2 -term first X 2 – 25x = -50, Divide each side by -2 X 2 – 25x + (-12.5) 2 = -50 + 156.25, Complete the square (x – 12.5) 2 = 106.25,Write as binomial squared X – 12.5 = + √106.25, Take the square roots of both sides X = 12.5 + √106.25,Solve for x X = 22.8 ft, Use a calculator to solve 50 – 45.6 = 4.4 So the garden is 22.8 by 4.4 ft. x 50 – 2x x

26. Discriminant In the quadratic formula, b 2 – 4ac under the radical sign is the discriminant which can tell you how many of what kind of solutions the equation has. If b 2 – 4ac > 0 the equation has 2 real solutions If b 2 – 4ac = 0 the equation has 1 real solution If b 2 – 4ac < 0 the equation has 2 imaginary solutions The standard form of a quadratic equation’s solution is: X = (-b + √b 2 – 4ac)/2a