After School review on Polynomials and Quadratics

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Quadratics and Polynomial Functions after School Review
1)
2) Convert to Vertex form and Identify the turning point (vertex)
a) π‘₯ 2 + 4π‘₯ + 2 = 0
b) βˆ’π‘₯ 2 + 6π‘₯ + 4 = 0
c) 4π‘₯ 2 + 40π‘₯ + 3 = 0
3) Solve by using square root property: 10π‘₯ 2 βˆ’ 10 = 630
4) Solve by factoring: 2π‘₯ 2 βˆ’ 11π‘₯ βˆ’ 21 = 0
5) Solve each equation using the quadratic formula: 6π‘₯ 2 + 8π‘₯ βˆ’ 25 = βˆ’3
6) Solve by completing the square: 3π‘₯ 2 + 18π‘₯ + 5 = 0
7) Write the center-radius form of the circle equation given the following.
A) πΆπ‘’π‘›π‘‘π‘’π‘Ÿ: (2, βˆ’6)π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  = 11
B) π‘π‘’π‘›π‘‘π‘’π‘Ÿ (βˆ’7, βˆ’3) π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  = 3√33
8) Convert π‘₯ 2 + 𝑦 2 βˆ’ 4π‘₯ βˆ’ 6𝑦 + 8 = 0 into center-radius form and identify its center and radius.
9) Find the quadratic equation given the following information.
a) πΉπ‘œπ‘π‘’π‘  (βˆ’3, 1), π·π‘–π‘Ÿπ‘’π‘π‘‘π‘Ÿπ‘–π‘₯ 𝑦 = 5
b) Vertex (3, 1), Focus (3, 5)
c) Vertex (5, -2), Directrix: y = -5
For this one, use the formula: x ο€­ h  ο€½ 4 p y ο€­ k 
2
10) Find the discriminant and state the number of solutions along with the description of the roots. (real,
imaginary, rational, irrational)
a) βˆ’6π‘₯ 2 βˆ’ 6 = βˆ’7π‘₯ βˆ’ 9
b) 4π‘₯ 2 + 5π‘₯ + 4 = βˆ’3π‘₯
11) Identify even, odd degree and positive, negative leading coefficient.
12) Is π‘₯ βˆ’ 5 a factor of π‘₯ 3 βˆ’ 7π‘₯ 2 + 2π‘₯ + 40. Justify your answer.
Use the remainder theorem.
Use long division
Find the other two factors.
13) Divide and state the quotient in quotient-remainder form.
14) Use this graph to answer the following questions.
A) What are the zeroes?
B) What are the factors?
C) Given the point (4, -32) what is the equation in standard form.
D) Put the equation in your calculator. Find the minimum and maximum.
E) Using the factored form of your equation, find the y-intercept.
Verify algebraically whether each function is even, odd, or neither!
1. 𝑓(π‘₯) = π‘₯ 3 βˆ’ 6π‘₯
2. 𝑔(π‘₯) = π‘₯ 4 βˆ’ 2π‘₯ 2
3. β„Ž(π‘₯) = π‘₯ 2 + 2π‘₯ + 1