Circle the Correct Answer s3

Grade 10 – LC3 – Reinforcement Lessons 5.1 to 5.4

Name: Date: ___/3/2010

Section: 10._____

Question I

Circle the correct answer

1. The minimum value of the Quadratic function : f ( x ) = x 2 – 4 x + 2 is

a. 0 b. 2 c. -2 d. -4

2. Use the graph to determine the solutions of the Quadratic function :

f(x) = x 2 – x – 6 a. -6 b. -2 c. -3 and 2 d. – 2 and 3

3. Simplify ( 2 – 5 i ) ( 2 + 5 i )

IAT-LC3 1/8 b. -1 c. 9 d. 4 – 10 i

4. Find a Quadratic equation with roots – 4 and 6

a. x2 – 24 x – 2 = 0 b. x 2 – 2 x – 24 = 0 c. x 2 + 2 x – 24 = 0 d. x 2 + 2 x + 24 =

IAT-LC3 2/8 Question II

f (x) = x 2 – 2 x - 4

1. Find the y-intercept, the equation of the axis of symmetry and the x coordinate of the vertex

y-intercept :

axis of symmetry :

vertex :

2. Make a table of values that includes the vertex

f ( x )

IAT-LC3 3/8 3. Use this information to graph this function

4. Use the related graph to determine its solutions

5. Determine whether the function has a maximum or a minimum value and find the maximum or minimum value

6. State the domain and the range of the function

IAT-LC3 4/8 Question III

Solve by Factoring :

1. x 2 – 4 x – 12 = 0

2. 9 x 2 – 36 = 0

3. x 2 – 9 x = 0

4. 2 x 2 – 12 x + 18 = 0

IAT-LC3 5/8 Question IV

Simplify :

1.  49 =

2.  25x 6 =

3.  64a 8b 6 =

4. ( 3 + 6 i ) ( 4 - 3 i ) =

5. ( 2i + 3 ) – ( 21i – 4 )

6. ( 2i + 5 ) + ( 32i - 6 )

7. (- 2i) (5 i) (-4 i) =

IAT-LC3 6/8 8. 2 i (- 2 i) 2 =

9. i 27=

2  i 10. 4  3i

11. Solve the equation below:

3 2 x - 1 = 0 4

12. Solve for m and n:

14 + 2i = 2m +9ni

IAT-LC3 7/8 13. The impedance in one part of a series circuit is 3 + 4j ohms, and the impedance in another part of the circuit is 2 – 6j. Add these complex numbers to find the total impedance in the circuit.

14. The voltage in a circuit is 14 – 8j volts, and the impedance is 2 – 3j ohms. What is the current?

IAT-LC3 8/8