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Name: Math 4
Review for Quarter 3 Cumulative Test
I. Solving quadratic equations Key Facts
• To factor a polynomial, first factor out any common factors, then use the box method to factor the quadratic.
o When you have a nonmonic quadratic ax2 + bx + c( ) , the numbers to fill in your box must multiply to a ⋅c and add to b
• To solve a quadratic equation, make sure one side of the equation is equal to 0. Then factor the quadratic and set all factors equal to 0. Solve all of the new equations for x.
• The quadratic formula gives the solutions to any quadratic ax2 + bx + c = 0
o x = −b± b2 − 4ac2a
Practice Problems 1. Factor the following expressions. Don’t forget to look for common factors first. a. 𝑥! − 3𝑥 − 10 b. 2𝑥! − 14𝑥! + 24𝑥 c. 4𝑥! − 8𝑥 − 5 d. 3𝑥! + 14𝑥 − 5 2. Solve the following equations by factoring and using the zero product property. a. 𝑥! − 4𝑥 − 12 = 0 b. 𝑥! + 3𝑥 = 0
3. Solve the following equations using the quadratic formula. Give your answers with square roots and as decimals. a. 𝑥! + 𝑥 − 4 = 0 b. 2𝑥! − 5𝑥 + 1 = 0 c. 3𝑥! − 𝑥 − 3 = 0 d. 𝑥! − 3𝑥 − 3 = 0 II. Finding the vertex of a parabola Key Facts
• Quadratic graphs have a parabola shape ∪ or∩( ) and are symmetric • There are two ways to find the vertex (min or max point) of a quadratic:
o Find the zeros (x-‐intercepts) by factoring or using the quadratic formula and average them to find the x-‐coordinate of the vertex
o Use the formula x = −b2a to find the x-‐coordinate of the vertex
• Once you know the x-‐coordinate of the vertex, plug that back into the original quadratic formula to find the y-‐coordinate. Remember, the vertex is a point so your answer should be in the form (x, y)
Practice Problems 4. Find the vertex of the following quadratic equations by solving and averaging the roots. a. 𝑥! + 12𝑥 + 32 = 0 b. 4𝑥! + 15𝑥 − 4 = 0
5. Find the vertex of the following quadratic equation by using the formula 𝑥 = !!!!
a. 𝑦 = 2𝑥! − 8𝑥 + 1 b. 𝑦 = 𝑥! + 3𝑥 − 10 III. Projectile Motion Key Facts
• Know how to use the formulas for projectile motion. The formulas will be given to you on the test, but you’ll need to know what the variables stand for:
Horizontal: x f = xi + vxit Vertical: yf = yi + vyit + 12 gt
2
vyf = vyi + gt
Practice Problems 6. A projectile is shot into the
air from a height of 5 meters. It is shot with an initial velocity of 30 m/s at an angle of 20°.
a. How far has the projectile traveled horizontally after 0.8 seconds?
Variable Meaning Variable Meaning xi
Initial horizontal position = 0 m
vxi Initial horizontal
velocity x f Final horizontal
position vyi Initial vertical
velocity
yi Initial vertical position
g Gravitational Constant = −10𝑚 𝑠!
yf Final vertical position t Time
Variable Known? Variable Known? xi vxi
x f vyi
yi g
yf t
Horizontal: x f = xi + vxit Vertical: yf = yi + vyit + 12 gt
2
vyf = vyi + gt b. How much time has passed when the ball hits the ground? Use the quadratic formula to solve. c. What is the maximum height the projectile reaches? At what time does it reach that height? This means you must find the vertex of a quadratic. 7. A projectile is shot into the air
from a height of 1 meter. It is shot with an initial velocity of 10 m/s at an angle of 70°.
a. What is the maximum height the projectile reaches? At what time does it reach that height?
b. What is the height of the projectile after 1.2 seconds?
Variable Known? Variable Known? xi vxi
x f vyi
yi g
yf t
c. How long does it take for the projectile to hit the ground? IV. Complex Numbers Key Facts
• To deal with negative numbers under square roots, we define the imaginary unit: i = −1 i2 = −1
o Example: −4 = 4 ⋅ −1 = 2i or −7 = 7 ⋅ −1 = 7i • A complex number (example 3− 4i ) is a number that has both a real part (3)
and an imaginary part (−4i ). • We add and subtract complex numbers by combining the real parts and the
imaginary parts separately (just like combining like terms). o Example: (1+ 5i)− (2−3i) = −1+8i
• We multiply complex numbers by completing a multiplication box and simplifying.
o This includes the extra step of substituting i2 = −1 ! • We divide complex numbers by multiplying the numerator and denominator
by the complex conjugate of the denominator. o To simplify the denominator, we can use the shortcut
(a+ bi)(a− bi) = a2 + b2 . o To simplify the numerator, complete a multiplication box.
Practice Problems 8. Simplify the following expressions. a. (3+ i)+ (2− 4i) b. (3+ i)− (2− 4i) c. (3+ i)(2− 4i) d. (1− i)(2i+ 6)
9. Simplify the following expressions. a. (2+ i)2 b. (3− i)(5+ i)− (2+3i)
c. 3+ i2+ 2i
d. 4+ 2i3− i
10. Solve the following equations using the quadratic formula. Write your answers in terms of i. a. x2 + 4x + 5= 0 b. 3x2 − 2x +1= 0
V. Graphing Equations and Inequalities (5.1-‐5.5) Key Facts
• Graphing one-‐variable inequalities (ex. x < 4 ) o Perform algebra steps to get x alone
If you multiply or divide by a negative number, you must flip the direction of the inequality
o Shade in the number line to represent solutions to the inequality Draw a closed circle if ≤ or ≥ Draw an open circle if < or >
• Graphing equations of lines o If the equation is in slope-‐intercept form ( y =mx + b ), m gives the
slope of the line (rise/run) and b gives the y-‐intercept o If the equation is in any form, you can plug in x = 0 to find the y-‐
intercept and plug in y = 0 to find the x-‐intercept • Graphing two-‐variable inequalities (ex. y > 2x −3)
o Graph the line as if it was an equation (=) Draw a dashed line if < or >
o Test a point to determine which half of the grid to shade • Solving systems of equations graphically
o Graph both lines o The point where they intersect is the solution to the system
If the lines are parallel, there is no solution If the lines are the same, there are infinitely many solutions
• Graphing a system of two-‐variable inequalities o Graph the solution to each inequality as described above o The final solution is where the shaded areas overlap (shade this
darker or with a new color) Practice Problems 11. Simplify and shade the solutions to each inequality on the number line provided. a. 5x + 2 <17
b. 6− 2x ≤14
12. Graph the solutions to the following inequalities on the grids below. Be sure to think about whether you need a solid or dashed line.
a. y ≥ 2x − 5 b. y < − 12 x +3
13. Solve the following systems of equations by graphing. Be neat and check your
solution by plugging into both equations if you are unsure.
a. y = x +1y = 3x − 5"#$
b. y = − 1
2 x + 24y+ 2x = 8"#$
Solution: Solution:
c. 2x −3y =12y = 2
3 x −1"#$
d. 3x = 9− yx − 2y = −4"#$
Solution: Solution: 14. Graph the solutions to the following systems of inequalities on the grids below.
Be neat and make sure it’s clear what the final solution is (shade darkly or use a different color).
a. y ≤ xy ≤ 2x − 6#$%
b. y > −3x + 4y ≤ − 2
3 x +1#$%
Answers 1. a. (𝑥 − 5)(𝑥 + 2) b. 2𝑥(𝑥 − 3)(𝑥 − 4) c. (2𝑥 + 1)(2𝑥 − 5) d. (3𝑥 − 1)(𝑥 + 5) 2. a. 𝑥 = 6,−2 b. 𝑥 = 0,−3 3. a. !!± !"
!≈ 1.56 and− 2.56 b. !± !"
!≈ 2.28 and 0.219
c. !± !"!
≈ 1.18 and− 0.847 d. !± !"!
≈ 3.79 and− 0.79 4. a. (−6,−4) b. (−1.875,−18.0625) 5. a. (2,−7) b. (−1.5,−12.25) 6. a. 22.55 meters b. 2.46 seconds c. height of 10.3 meters after 1.03 seconds 7. a. height of 5.42 meters after 0.94 seconds b. 5.08 meters c. 1.98 seconds 8. a. 5−3i b. 1+ 5i c. 10−10i d. 8− 4i
9. a. 3+ 4i b. 14− 5i c. 8− 4i8
=1− 12 i d. 10+10i
10=1+ i
10. a. x = −4± 2i2
= −2± i b. x = 2± i 8
6
11. a. b. 12. a. b.
13. a. (3, 4) b. infinitely many solutions c. no solution d. (2, 3) 14. a. b.