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Math 130 College Algebra

  • If you are considering taking a lower-level class than Math 130:
  • If you are considering taking a higher-level class than Math 130:
 
  • Sample of what you should know before taking Math 130:
    1. Solve open parenthesis x squared minus 2 close parenthesis squared minus open parenthesis x squared minus 2 close parenthesis minus 6 equals 0
    2. If f of x equals x squared minus 2 x , find f of open parenthesis a plus h close parenthesis
    3. Graph Y equals negative open parenthesis x plus 2 close parenthesis squared plus 4.  Label vertex and all intercepts.
    4. Solve start set top row negative x plus 2 y minus z equals 1 second row negative x minus y minus z equals negative 2 bottom row x minus y plus 2 z equals 2 end set  using Elimination
    5. Solve 3 x squared equals 8 x minus 7
    6. log base 3 of open parenthesis x minus 5 close parenthesis plus log base 3 of open parenthesis x plus 3 close parenthesis equals 2
    7. A model rocket is launched upward from a platform 40 feet above the ground.  The quadratic function  s of t equals negative 16 t squared plus 400 t plus 40  models the rocket’s height above the ground, s of t, in feet, seconds after it was launched.  What is the maximum height that the rocket will reach?
      • Solution
        t equals start fraction negative b over 2 a end fraction equals start fraction negative 400 over 2 times negative 16 end fraction equals 12 point 5 seconds
        s of 12 point 5 equals negative 16 times 12 point 5 squared plus 400 times 12 point 5 plus 40 equals 2450 feet
        The rocket will reach a maximum height of 2,540 feet above the ground.
     
  • Sample of what you will learn in Math 130:
    1. Prove 5 plus 10 plus 15 plus dot dot dot plus 5n equals start fraction 5 n open parenthesis n plus 1 close parenthesis over 2 end fraction
      • Solution
        Proof:  If n equals 1 then 5 times 1 equals start fraction 5 open parenthesis 1 plus 1 close parenthesis over 2 end fraction
        5  =  5  so statement is true.
         
        Suppose the statement is true for n = k, then

        5 plus 10 plus 15 plus dot dot dot plus 5 k equals start fraction 5 k open parenthesis k plus 1 close parenthesis over 2 end fraction

        5 plus 10 plus 15 plus dot dot dot plus 5 k plus 5 open parenthesis k plus 1 end parenthesis equals start fraction 5 k open parenthesis k plus 1 close parenthesis over 2 end fraction plus 5 open parenthesis k plus 1 end parenthesis

         5 plus 10 plus 15 plus dot dot dot plus 5 k plus 5 open parenthesis k plus 1 end parenthesis equals start fraction 5 k open parenthesis k plus 1 close parenthesis over 2 end fraction plus start fraction 10 open parenthesis k plus 1 close parenthesis over 2 end fraction

        5 plus 10 plus 15 plus dot dot dot plus 5 k plus 5 open parenthesis k plus 1 end parenthesis equals start fraction 5 k open parenthesis k plus 1 close parenthesis plus 10 open parenthesis k plus 1 close parenthesis over 2 end fraction

        5 plus 10 plus 15 plus dot dot dot plus 5 k plus 5 open parenthesis k plus 1 end parenthesis equals start fraction 5 open parenthesis k plus 1 close parenthesis open parenthesis k plus 2 close parenthesis over 2 end fraction

        10 plus 15 plus dot dot dot plus 5 k plus 5 open parenthesis k plus 1 end parenthesis equals start fraction 5 open parenthesis k plus 1 close parenthesis open parenthesis open parenthesis k plus 1 close parenthesis plus 1 close parenthesis over 2 end fraction

        Hence, the statement is true for n = k + 1.
        Therefore, the statement is true for all natural numbers.
    2. Find a 1 for the arithmetic sequence given S 16 = – 160  and a 16 = – 25  
    3. Graph F of x equals start set top row x plus 5 with x less than negative 2 second row x squared with negative 2 less than or equal to x less than 1 bottom row square root of x with x greater than or equal to 1
    4. Solve open set top row negative x plus 2 y minus z equals 1 second row negative x minus y minus z equals negative 2 bottom row x minus y plus 2 z equals 2 close set  using Gauss-Jordan elimination with matrices
    5. Perform each operation when possible: 
    a. start 3 by 2 matrix top row 1 2 second row 3 0 bottom row negative 6 5 end matrix times start 2 by 2 matrix top row 4 8 bottom row negative 1 2 end matrix  
    b. start 2 by 2 matrix top row 4 8 bottom row negative 1 2 end matrix times start 3 by 2 matrix top row 1 2 second row 3 0 bottom row negative 6 5 end matrix
    c. start 2 by 3 matrix top row 2 5 8 bottom row 1 9 2 end matrix minus start 2 by 3 matrix top row 4 8 0 bottom row negative 1 2 negative 3 end matrix
      • Solution
        a. 3 by 2 matrix top row 2 12 second row 12 24 bottom row negative 29 negative 38
        b.  cannot be multiplied
        c. 2 by 3 matrix top row negative 2 negative 3 8 bottom row 2 7 5
    6. Graph  g of x equals start fraction x squared minus 2 x minus 3 over 2 x squared minus x minus 10 end fraction    identifying any intercepts and asymptotes
      • Solution
        x – intercepts: (3, 0), (– 1, 0)
        y – intercept: (0, 0.3)
         vertical asymptotes: x = 2.5,  x = – 2
         horizontal asymptote: y = 0.5
         graph crosses asymptote at
        graph with x scale from negative 8 to 8 and y scale from negative 8 to 8 comma vertical asymptotes at x equals negative 2 and x equals 2 point 5 and horizontal asymptote at y equals point 5 comma from left to right the graph starts close to y equals point 5 but above it and curves upward towards a y value of infinity as the x values get close to negative 2 comma on the right side of x equals negative 2 the graph starts close to a y value of negative infinity and curves upward crossing the y axis at point 3 and the x axis at negative 1 and then continuing up towards a y value of infinity as the x values get close to 2 point 5 comma on the right side of x equals 2 point 5 the graph starts close to a y value of negative infinity and curves upward towards a y value of point 5 as x approaches infinity
    7. Solve  2 e superscript negative 2 x baseline minus e superscript negative x baseline equals 1
    8. Find all zeros of the function:  h of x equals 6 x cubed plus 17 x squared minus 31 x minus 12
      • Solution

        The possible rational roots are: ±1, 2, 3, 4, 6, 12, ½, 3/2, 1/3, 2/3, 4/3, 1/6 

        Synthetic division table with root 1 in the upper left corner with a horizontal line under the 1 and a vertical line to the right of the 1 comma top row 6 17 negative 31 negative 12 comma second row blank space 6 23 negative 8 comma a horizontal line to separate the second and third rows comma third row 6 23 negative 8 negative 20

        Synthetic division table with root negative 1 in the upper left corner with a horizontal line under the negative 1 and a vertical line to the right of the negative 1 comma top row 6 17 negative 31 negative 12 comma second row blank space negative 6 negative 11 42 comma a horizontal line to separate the second and third rows comma third row 6 11 negative 42 30

        Synthetic division table with root 2 in the upper left corner with a horizontal line under the 2 and a vertical line to the right of the 2 comma top row 6 17 negative 31 negative 12 comma second row blank space 12 58 54 a horizontal line to separate the second and third rows comma third row 6 29 27 42

        Synthetic division table with root negative 2 in the upper left corner with a horizontal line under the negative 2 and a vertical line to the right of the negative 2 comma top row 6 17 negative 31 negative 12 comma second row blank space negative 12 negative 10 82 comma a horizontal line to separate the second and third rows comma third row 6 5 negative 41 70

        Synthetic division table with root negative 3 in the upper left corner with a horizontal line under the negative 3 and a vertical line to the right of the negative 3 comma top row 6 17 negative 31 negative 12 comma second row blank space negative 18 3 84 a horizontal line to separate the second and third rows comma third row 6 negative 1 negative 28 72

        Synthetic division table with root negative 4 in the upper left corner with a horizontal line under the negative 4 and a vertical line to the right of the negative 4 comma top row 6 17 negative 31 negative 12 comma second row blank space negative 24 28 12 comma a horizontal line to separate the second and third rows comma third row 6 negative 7 negative 3 0

        open parenthesis x plus 4 close parenthesis open parenthesis x squared minus 7 x minus 3 close parenthesis equals 0

        x squared minus 7 x minus 3 equals 0

        x equals start fraction 7 plus or minus start square root 49 minus 4 times negative 3 end square root over 2 end fraction equals start fraction 7 plus or minus start square root 61 end square root over 2 end fraction

        Zeros of h(x) are:  negative 4 and start fraction 7 plus or minus start square root 61 end square root over 2 end fraction