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Unit 3: Polynomials Math 2 (Fall 2015) Name: ______________________________ Date: ______________ Period _______ Test Review: Due on Test day BEFORE the test. Graded for ACCURACY! Every day late is minus 5 points. Show all work. You will not receive full credit if you do not show your work. If I cannot read the work then it will be counted wrong. NO Homework passes accepted for this assignment. *This is not all inclusive. Please study all quizzes and warm ups. You need to know & be able to do Things to remember Vertex: Point where the graph changes direction. Practice Problems For the following problems find the characteristics 1) . 2) 𝑦 = (𝑥 − 2)2 + 3 Vertex: Maximum: The parabola opens down (if 𝒂 is negative) Minimum: The parabola opens up (if 𝒂 is positive) Axis of Symmetry: The vertical line cutting the parabola in half Find the Characteristics of a parabola y-intercept: The point where the parabola crosses the y-axis Vertex Form 𝑦 = 𝒂(𝑥 − 𝒉)2 + 𝒌 Vertex: (h, k) AoS: x = h y-intercept: (0, #), plug 0 in for x Standard Form 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 Vertex: (#, #) 𝑏 𝒙=− 2𝑎 𝒚 =plug x into equation 𝑏 AoS: 𝒙 = − 2𝑎 y-intercept: (0, #), plug 0 in for x Max or Min: AoS: y-int: Vertex: Max or Min: AoS: y-int: 3) 𝑦 = 𝑥 2 − 8𝑥 + 7 Vertex: 4) 𝑦 = −3(𝑥 + 1)2 − 2 Vertex: Max or Min: Max or Min: AoS: AoS: y-int: y-int: Graph a quadratic function 1. Find and plot the vertex. State the characteristics and graph the following equations. 5) 𝑦 = (𝑥 − 3)2 − 9 2. Find and graph the axis of symmetry (dashed vertical line) Vertex: 3. Find the points on the each side of the vertex and plot. (calculator can help you with this) AoS: Max or Min: y-int: Solutions: 6) 𝑦 = −𝑥 2 + 6𝑥 − 5 Vertex: Max or Min: AoS: y-int: Solutions: Solve a Quadratic by Factoring Get in standard Form Solve by factoring 7) 9𝑥 2 − 25 = 0 8) 𝑥 2 = 9𝑥 − 14 9) 𝑥 2 = −4𝑥 + 12 10) 2𝑥 2 − 20 = 6𝑥 Factor Set each factor equal to zero and solve Find the discriminate and state the number and type of solutions. 11) 2𝑥 2 + 7𝑥 + 3 = 0 12) 𝑥 2 + 4𝑥 − 1 = −5 𝑥= −𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 2 Solve a Quadratic by Quadratic Formula Discriminate: 𝑏 − 4𝑎𝑐 Positive answer 2 real roots. Zero, one real root (two equal real roots). Negative answer 2 imaginary roots. To Solve: 1st: Put into standard form. nd 2 : Plug a, b, and c into the quadratic formula 3rd: type the discriminate in the calculator 4th: Simplify the radical 5th: simplify the numbers 13. −4𝑥 2 + 4𝑥 = 5 14. −𝑥 2 = 4𝑥 Find the solutions by using the quadratic formula. 13. 3𝑥 2 = −10𝑥 − 3 14. 3𝑥 2 + 3𝑥 = 1 15. 𝑥 2 + 4𝑥 − 2 = 0 16. 7𝑥 2 = 14 + 2𝑥 Solve the equation using the square root method. 17. 𝑥 2 = 20 18. 𝑥 2 − 16 = 0 19. 3𝑥 2 − 12 = 0 20. 9𝑥 2 − 25 = 0 Isolate the square. Solve a Quadratic by Taking Square Root Take the square root of both sides. Don’t forget the ± Simplify. Transform a quadratic functions Parent Function 𝑦 = 𝑥2 Describe the transformations 21. 𝑦 = 𝑥 2 − 3 22. 𝑦 = (𝑥 − 3)2 Transformations 𝑦 = 𝑥 2 + 𝑘 vertical shift +k up and –k down 𝑦 = (𝑥 + ℎ)2 horizontal shift +h left and –h right 𝑦 = −𝑥 2 reflect graph over the x-axis 23. 𝑦 = 3 𝑥 2 1 24. 𝑦 = −𝑥 2 25. 𝑦 = 4(𝑥 + 2)2 − 5 26. 𝑦 = − 2 𝑥 2 + 4 3 Write the equation for the following transformations 𝑦 = 𝑎𝑥 2 |𝑎| > 1 compressed (wide) 0 < |𝑎| < 1 stretched (wide) 27. The graph goes down 7 and right 3 Convert from Vertex Form to Standard Form 1. Foil or the square term. 2. Distribute the “a” value. 3. Combine like terms. Write in standard form 5. 𝑦 = (𝑥 − 3)2 + 1 Convert from Standard Form to Vertex Form Write in vertex form 1. Find the vertex 2. Identify the “a” value. 7. 𝑦 = 2𝑥 2 − 4𝑥 − 1 3. Use these three values and substitute into the form: 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 28. The parabola is flipped and compressed by a factor of 1/4. Its position moves left 4 and up 8 6. 𝑦 = −3(𝑥 + 2)2 − 4 8. 𝑦 = −𝑥 2 − 8𝑥 + 4