Breaking a quadrilateral down into its coefficient pieces: y = ax 2 + bx + c

1. Breaking a quadrilateral down into its coefficient pieces: y = ax2 + bx + c • a > 0 (+) parabola opens up; parabola has a min value • a < 0 (-) parabola opens down parabola has a max value • c is the y-intercept of the parabola x-intercepts are the zeros (see them later) • Vertex formula: (x, y) =( -b / 2a , (b2 – 4ac)/(4a) ) (vertex y-value is the min or the max) • (Use max or min part of 2nd TRACE in calculator to let the calculator find it) • Line of Symmetry: x = -b/2a • Domain: X is all real numbers Range: Y is ≥ min if it opens up (a > 0) Y is ≤ max if it opens down (a < 0) • Find for y = -x2 - 5x + 14 a. Opens: Up Down • Domain: {x | • Range: {y | • Vertex: ( , ) • Line of Symmetry: x = • Y-intercept: (0, ) • Find for y = x2 + 13x + 40 a. Opens: Up Down • Domain: {x | • Range: {y | • Vertex: ( , ) • Line of Symmetry: x = • Y-intercept: (0, ) • Find for y = -2x2 + 2x + 40 a. Opens: Up Down • Domain: {x | • Range: {y | • Vertex: ( , ) • Line of Symmetry: x = • Y-intercept: (0, ) • Find for y = x2 – 6x + 9 a. Opens: Up Down • Domain: {x | • Range: {y | • Vertex: ( , ) • Line of Symmetry: x = • Y-intercept: (0, ) 28

2. Finding zeros of polynomials and solving equations = 0 are the same thing. They can be found be graphing the polynomial and finding the x-intercepts (2nd Trace, Zeros) • Examples: • f(x) = x(x + 4) (x – 1) 0 = (2x – 1)(3x + 2) • zeros of f(x) are when x = 0, x = -4 and x = 1 zeros of g(x) are when x = ½ • and x = -2/3 • x = 0, (x+4) = 0, (x – 1) = 0 (2x – 1) = 0, (3x + 2) = 0 • x = -4 x = 1 2x = 1 3x = -2 • x = ½ x = -2/3 • Find the zeros of • f(x) = (x – 2) (x + 3) • g(x) = (x + 6)(x – 1) • h(x) = (2x + 1)( x – 7) • j(x) = (3x + 4)(2x – 5) • f(t) = (t – 6)((t – 1)(t + 3) • Find the solutions to • (x + 2) (x + 5) = 0 • (h – 2)(2h – 3) = 0 • (x + 4)(x – 1) = 0 • (m – 3)( 2m – 7) = 0 • t(t + 4)(t – 8) = 0 22

3. Quadratic Regression: y = ax2 + bx + c Put x-values in L1 and y-values in L2. Go to STAT – CALC and select 5: QuadReg Write down values calculator gives for a, b and c for quadratic equation above. Use new equation to predict values inside bounds of data (interpolation) or to predict values beyond the bounds of the data (extrapolation). Graph data using StatPlot (Zoom 9 sets up the window properly) and graph the regression model using Y1 = ax2 + bx + c Write the function: y = __________________________________ What would the model predict if x = 5? • Quadratic Equation: Must be in the form of ax2 + bx + c = 0 • -b  √b2 – 4ac • x-intercepts (zeros of the function) = ----------------------------- • 2a • Remember a negative number squared is now a positive number! • Using the Quadratic Equation on the following: • x2 + 5x – 14 = 0 • 2x2 + 13x + 21 =0 • 6x2 + 5x – 5 = 0 • x2 + 3 = 0What does getting a negative number under the radical (in #4) mean on the graph of the problem? 10

4. Factoring: • Factor Cases: Looking at the signs • 1) + + + = ( + )( + ) • 2) + - + = ( - )( - ) • 3) + - - = ( + ) ( - ) where the |negative| > |positive| • 4) + + - = ( + ) ( - ) where the |positive| > |negative| • Example 1: x² - 5x + 6 = 0 identify coefficients: a = 1, b = -5, c = 6 • If a ≠ 1, then break a down into its factors or factor a out of each term • Case 2 applies. x² - 5x + 6 = (x - )(x - ) • Break down c into its factors: 16 and 23 1, 2, 3, 6 • If a = 1 then what combinations of the factors of c add up to b -5 = -2 + -3 • So solution is x² - 5x + 6 = (x - 2)(x - 3) Zeros are at x = 2 and x = 3 • Factor the following problems: • y = x2 + 5x + 6 • y = x2 + 15x + 56 • y = x2 - 7x + 12 • y = x2 + 2x + 1 • y = x2 - 5x - 14 • y = x2 - 6x + 9 • y = x2 + x - 6 • y = x2 - 12x + 32 16