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The purpose of precalculus is to prepare students for calculus, regardless of whether they go on to take it. In preparing for college level math, students must solidify and expand their understanding of previously learned math topics. This summer assignment reviews topics from algebra and geometry that are necessary to be successful in precalculus. Although this assignment will not be collected and graded, you will be given a summative assessment on this material during the second day of class. Remember that you must pass math your senior year to graduate, so if you are a senior and find yourself struggling to complete this assignment, please contact your counselor and switch courses prior to August 1st . This assignment and the solutions will be located on the school website under the information tab 2016 summer school. If you have trouble accessing the assignment contact student services. It is also highly recommended that you use google and youtube to search for the topic you need assistance in. Here are two sites that offer help. http://www.purplemath.com/modules/index.htm http://khanacademy.org
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Page 1: The purpose of precalculus is to prepare students for ... · PDF fileThe purpose of precalculus is to prepare students for calculus, regardless ... This summer assignment reviews topics

The purpose of precalculus is to prepare students for calculus, regardless of whether they go on to take it. In preparing for college level math, students must solidify and expand their understanding of previously learned math topics.

This summer assignment reviews topics from algebra and geometry that are necessary to be successful in precalculus. Although this assignment will not be collected and graded, you will be given a summative assessment on this material during the second day of class.

Remember that you must pass math your senior year to graduate, so if you are a senior and find yourself struggling to complete this assignment, please contact your counselor and switch courses prior to August 1st.

This assignment and the solutions will be located on the school website under the information tab 2016 summer school. If you have trouble accessing the assignment contact student services. It is also highly recommended that you use google and youtube to search for the topic you need assistance in. Here are two sites that offer help.

http://www.purplemath.com/modules/index.htmhttp://khanacademy.org

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A. FOIL  -­‐   mul,ply  binomials      first  -­‐  outside  -­‐  inside  -­‐  last

(x  +  2)(x  -­‐  1)  = (x  -­‐  4)(x  +  3)  =(4x2  -­‐  3)(2x  +  7)  =   (6  -­‐  m3)(2  +  m)  =(3x  +  y)(m  -­‐  5y)  =   (x  +  y)(x  -­‐  y)  =                                                                          (3  -­‐  m)(m  +  m2)  = (5  -­‐  x)(3  -­‐  x)  =

 

 

B. Factoring  -­‐      ax2  +  bx  +  c          when  a  =  1    find  two  numbers  that  mul,ply  to  make  "c"  and  add  to  make  "b"  

x2  +  14x  +  48  =  (                )(              ) b2  -­‐  8b  +  15  =  (                )(              )p2  +  14p  +40  =   b2  -­‐  9b  +  14  =

when  a  ≠  12x2  +  12x  -­‐  14  = 4x2  +  8x  +  3  =3b2  -­‐  3b  -­‐  36  = 5a2  -­‐  23a  +  12  =

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C. Frac,ons  -­‐  Adding/Subtrac,ng  (get  a  common  denominator)

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D. Frac,ons  -­‐  Mul,plying/Dividing  

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E. Quadra,c  Formula  (solve  a  quadra,c  to  find  the  zeros)

x2  -­‐  5x  +  4  =  0 x2  -­‐  2x  -­‐  24  =  0x2  +  6x  =  -­‐9 x2  -­‐  10  =  -­‐3x2x2  -­‐  x  -­‐  6  =  0  -­‐  x  +  4  =  2x2

F. Solve  each  equa,on  for  the  given  variable

2x  +  1  =  5x  -­‐  2 7m  +  2  =  37x2  +  4  =  29 32w  +  ½  =  16.52x2  -­‐  98  =  0   -­‐  x  +  4  =  22  +  x

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G. Write  an  equa,on  of  a  line  according  to  the  given  informa,on

containing  (-­‐1,  -­‐4)  and  parallel  to  y  =  3x  +  2

containing  (2,  -­‐4)  and  parallel  to  x  -­‐  2y  =  5

containing  (-­‐2,  3)  and  parallel  to  x  =  1

containing  (4,  15)  and  parallel  to  -­‐x  +  2/3  y  =  6

containing  (2,  3)  and  perpendicular  to  y  =  2x  -­‐  1

containing  (1,  -­‐3)  and  perpendicular  to  y  =  -­‐  3

containing  (3,  4)  and  perpendicular  to  2x  -­‐  3y  =  -­‐6

containing  (4,  1)  and  perpendicular  to  ½x  +  y  =  3

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H.Use  the  exponent  rules  to  simplify  each  expression.All  answers  should  be  wriben  with  posi,ve  exponents.

(m3)4  =   (3c6)2  =  (m3)(m4)  =   (5c6)2(2c)  =  (5a2)(3a3)  =   (-­‐7cd2)(3c-­‐2)  =(-­‐s3t)(-­‐5t4)  =   (m3n2)(4m2n-­‐2)  =(3a2b4c)(7a3b3)  =   (-­‐2cd2)2(3c2)3  =a0  = m-­‐1  =

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I. Find  each  sum  or  difference.    Express  all  answers  in  standard  form.

(x2  +  3x  +  2)  +  (3x2  +  x  -­‐  6)  =

(x2  +  3x  +  2)  -­‐  (3x2  +  x  -­‐  6)  =

(3a4  -­‐  2a2  -­‐  1)  +  (2a3  +  2a2  -­‐  10)  =  

(3a4  -­‐  2a2  -­‐  1)  -­‐  (2a3  +  2a2  -­‐  10)  =  

(-­‐  2a2  -­‐  1)  +  2(3b2  -­‐  5)  =  

(3x  +  4xy  -­‐  7y)  +  (-­‐x  -­‐2xy  +  4y)  =

(-­‐5x2  -­‐  2x  +  1)  -­‐  (3x2  +  4x  -­‐  2)  -­‐  (-­‐8x2  -­‐  5x  -­‐  3)  =  

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Appendix A.2 Polynomials and Factoring

A polynomial is any expressions that can be written:

anxn + an-1xn-1 + ... + a1x + a0

(where n is a nonnegative integer and an ≠ 0)

Standard Form: exponents are in descending order

Degree: highest exponent (all exponents must be +)Addition: combine like terms

Subtraction: change to addition and combine like termsMultiply Binomials: FOIL

Completely Factored:written as a product of its primes

(5x-7)+(2x2+10x) =2x2 + 15x - 7

(10m-3) - (-4m - 2) = 14m - 1

(x3-x2)(x2 + 2) = x5 +2x3 - x4 - 2x2

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Steps for Factoring (Grouping-Method)

1.) First, write the equation in Standard Form!

y = ax + bx + c

2.) Next, label a, b, and c.

3.) Multiply a*c

4.) List the factors of a*c

5.) Replace the "b" term with the factor pair of

a*c that adds to get "b".

6.) Group the first two terms together, and the second two terms together

7.) Pull out the GCF

8.) Write the two binomial answers.

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ax2 + bx + c

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fractions

When adding/subtracting find a common denominator

1 + 23 5

55

33

515 + 6

15

1115

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fractions

multiply: numerator x numeratordenominator x denominator

2 35 7

= 635

divide: multiply by the reciprocal of the denominator

2 35 7 = 2 7

5 3= 14

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