Graphical Methods Questions and Answers - Form 3 Topical Mathematics

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Questions

  1. The equation of a circle is given by x2 + 4x + y2 – 5 = 0. Find the centre of the circle and its radius.
  2. The equation of a circle is x2 + y2 + 6x – 10y – 2 = 0. Determine the co-ordinates of the centre of the circle and state its radius.
  3. In the diagram below ABE is a tangent to a circle at B and DCE is a straight line.
    If ABD = 60
    o, BOC = 80o and O is the centre of the circle, find with reasons ∠BEC
  4. Obtain the centre and the radius of the circle represented by the equation:
    x2 + y2 – 10y + 16 = 0.
  5. Complete the table below, for the function y = x3 + 6x2 + 8x
    -5 -4 -3 -2 -1 0 1
    x3 -125   -27 -8   0 1
    6x2   96 54   6 0 6
    8x -40   -24     0 8
    y     3 0   0 15
    1. Draw a graph of the function y = x3 + 6x2 + 8x for – 5 ≤ x ≤ 1 and use the graph to estimate the roots of the equation x3 + 6x2 + 8x = 0.
    2. Find which values of x satisfy the inequality x3 + 6x2 + 8x -1 > 0
  6. Sketch the curve of the function y = x3– 3x + 2 showing clearly minimum and maximum points and the y – intercept.
  7. Show that 4y2 + 4x2 = 12x – 12y + 7 is the equation of a circle, hence find the co-ordinates of the centre and the radius.
  8. Two variables R and P are connected by a function R = KPn where K and n are constants. The table below shows data involving the two variables.
    3 3.5 4 4.5 5
    36 49 64 81 100
    1. Express R = KPn in a linear form
    2. Draw a line graph to represent the information above
    3. Find the values of constants K and n
    4. Write down the law connecting R and P
    5. Find the value of P when R = 900
  9. A circle of radius 3cm has the centre at (-2, 3) . Find the equation of the circle in the form of x2 + y2 + Px + qy + c = 0
  10. In an experiment, the values of two quantities V and T were observed and the results recorded as shown below.
    V 0 2 4 6 8 10
    T 0.49 0.30 0.24 0.20 0.16 0.137

    It is known that T and V are related by a law of the form
    T =    a    
          b + V
    where a and b are constants.
    1. Draw the graph of I/T against V
    2. Use your graph to find;
      1. The values of a and b.
      2. V when T = 0.38
      3. T when V = 4.5
  11. Find the equation of the tangent to the curve y = 2x3 + x2 + 3x – 1 at the point (1, -5) expressing you answer in the form y = mx + c.
  12. Given that :- 243 = (81)-1 x ( 1/27)x determine the value of x
  13. Show that 3x2 + 3y2 + 6x – 12y - 12 = 0 is an equation of a circle hence state the radius and centre of the circle.
    1. Fill in the table below for the function y = -6 + x + 4x2 + x3 for -4 x 2
      x -4 -3 -2 -1 0 1 2
      -6 -6 -6 -6 -6 -6 -6 -6
      x -4 -3 -2 -1 0 1 2
      4x2               
       x3              
       y              
    2. Using the grid provided draw the graph for y = -6 + x + 4x2 + x3 for -4≤x ≤2
    3. Use the graph to solve the equations:-
      1. x3 + 4x2 + x – 4 = 0
      2. -6 + x + 4x2 + x3 = 0
      3. -2 + 4x2 + x3 = 0
  14. The table below shows the results obtained from an experiment to determine the relationship
    between the length of a given side of a plane figure and its perimeter
    Length of side l (cm) 1 2 3 4 5
    Perimeter P(cm) 6.28 12.57 18.86 21.14 31.43
    1. On the grid provided, draw a graph of perimeter P, against l
    2. Using your graph determine;
      1. the perimeter of a similar figure of side 2.5cm
      2. the length of a similar figure whose perimeter is 9.43cm
      3. the law connecting perimeter p and the length l
    3. If the law is of the form P = 2k+ c where k and c are constants, find the value of k
  15. In an experiment with tungsten filament lamp, the reading below of voltage (V) current (I), power (P) and resistance (R)were obtained. It was established that P was related to R by a law P = a Rn – 0.6. Where a and n are constants.
    V 1.30 2.00 2.80 4.40 5.70
    I 1.50 1.80 2.10 2.50 2.90
    P 0.73 2.05 3.28 7.44 10.62
    R 0.89 1.13 1.33 1.78 1.99
    Plot a suitable line graph and hence use it to determine the value of a and n
  16. Find the gradient of a line joining the centre of a circle whose equation is x2 + y2 – 6x = 3 – 4y and a point P(6,7) outside the circle..
    1. Complete the table below for the function y = -x3 + 2x2 – 4x + 2.
      x -3 -2 -1 0 1 2 3 4
      -x3 27 8   0   -8    
      2x2 18 8 2 0        
      -4x   8   0        
      2 2 2 2 2 2 2 2 2
      y   26   2   -6   -46
    2. On the grid provided below draw the graph of -x3 + 2x2 – 4x + 2 for - 3 x 4.
    3. Use the graph to solve the equation -x3 + 2x2 – 4x + 2 = 0.
    4. By drawing a suitable line on the graph solve the equation. –x3 + 2x2 – 5x + 3 = 0.
  17. Determine the turning point of the curve y = 4x3 - 12x + 1. State whether the turning point is a maximum or a minimum point.
    1. Complete the table below for the equation of the curve given by y = 2x3 – 3x2 + 1
      X -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
      2x3 -16   -2   0   2   16    
      -3x2 -12     0.75 0 0.75         -27
      1 1       1            
      y -27 -12.5     1           13.5
    2. Use the table to draw the graph of the function y = 2x3 – 3x2 + 1
    3. Use your graph to find the values of x for :-
      1. y > 0
      2. The roots of the equation 2x3 – 3x2 + 1 = 0
      3.  2x3 – 3x2 = 9
  18. Find the radius and the centre of a circle whose equation is :
    2x
    2 + 2y2 – 6x + 10y + 9 = 0

Answers

  1. x2 + 4x + y2 = 5
    x2 + 4x + ( ½ x 4)2 + y2 = 5 + (½ x 4)2
    (x + 2)2 + (y + 0)2 = 5 + 4
    (x + 2)2 + (y + 0)2 = 9
    Centre (-2,0)
    Radius √9
    r = 3 units
  2.  x2 + 6x + (3)2 + y2 – 10y + (-5) = 2 + 9 + 25
    (x + 3)2 + (y – 5)2 = 36
    (x – -3)2 + (y - +5)2 = 62
    centre (-3, 5)
    Radius 6 units
    Completing of sq. for expression in x and y.
    √ Expression.
    Centre
    Radius
  3. CBE = 40o ( alt.segiment theoren)
    ∠BCE = 120o (Suppl. To BCD = 600 alt. seg.)
    ∠(40 + 120 + E) = 180o (Angle sum of Δ)
    BEC = 20o
  4. X2 +Y2 – 10Y + 25 = 25 – 16
    (X -0)2 + (Y – 5)2 = 9
    (X – 0)2 + (Y – 5)2 = 32
    Centre (0, 5)
    Radius = 3
  5.  
    x -5 -4 -3 -2 -1  1
    x3 -125  -64  -27  -8 -1  1
    6x2  150 96 54 24 6 6
    8x -40 -32  -24  -16  -8  8
    y -15 0 3 0 -3  15
    x3+ 6x2 + 8x >1
    Between
    1. x = -3.85 ± 0.1 and x = - 2.15± 0.1
    2. x > 0.5 ± 0.1
  6. y = x3 – 3x + 2
    x = 0, y = 2
    (0, 2) y – intercept.
    dy = 3x2 – 3 = 0
    dx     x2 = 1
    x = 1
    x = 1 y = 0
    Point (1, 0) min point
    x = -1, y= 4
    Point (-1, 4) max point
    graphical ans6
  7. 4x2 – 12x + 4y2 + 12y = 7
    x2– 3x + y2 + 3y = 7/4
    x2– 3x + (3/2)2 + y2 + 3y + (3/2)2 = 7/4 + 9/4 + 9/4 = 25/4
    (x – 3/2)2 + (y + 3/2)2 = 25/4
    ∴Centre (1,5, -1.5) Radius 2.5units
  8. Log R =nlog p + log K
    Log P  0.48  0.54  0.60  0.65  0.70
    Log R  1.56  1.69  1.81  1.91  2.00
    Gradient = 2 – 0.6
                       0.7
    = 1.4 = 2
       0.7
    Log R intercepts = 0.6 = logk
    K= 4
    The law connecting R and P is R=4P2
    900 = 4P2
    P2 = 900
             4
    225 = P2
  9. (x +2)2 (y-3)2 = 32
    X2 + 4x + 4 + y2 – 6y + 9 = 32
    X2 + y2 + 4x – 6y + 4 = 0
  10.  
    1.  
      V 0 2 4 8 10
      1 T 2.04  3.33  4.17  6.25  7.30
         T     = a
      b + V
       I  = b + V
      T        a
       I  = 1V + b
      T      a      a
      y = mx + C
      1. 1 = Grad y = 7.3 – 5 = 2.3 = 0.575
        a                x    10 – 6      4
        a = 1.739
        b = y – Intercept 2.04
        a
           b    = 2.04  b = 2.04 x 1.739
        1.739               = 3.547556
        b 3.548
      2. T = 0.38
         I   = 2.63 shown on graph
        T
        V = 1
        -1
      3.  I  = 4.45
        T
        T = (4.45)
        = 0.2247
        0.22
  11. y = 2x3 + x2 + 3x -1
    dy = 6x2 + 2x + 3
    dx
    gradient at (1, -5)
    = 6 + 2 + 3= 11
    y-(-5) = 11
    x - 1
    y + 5 =11x -11
    y = 11x -16
  12. 35 = 3-4 x 3-x
    35 = 3-4-x
    -4 –x = 5
    -x = 9
    x =-9
  13.  x2 + 2x + 1 + y2 – 4y + 4 = 4 + 1 + 1
    (x+1)2 + (y-2)2 = 9
    Centre (-1, 2)
    Radius 3 units
    1.  
      X -4 -3 -2  -1  0 1 2
      -6 -6 -6 -6  -6  -6  -6  -6
      X -4 -3 -2  -1  0 1 2
      4x2 64 36 16  4 0 4 16
      X3 -64  -27  -8  -1  0 1 8
      Y=-6+x+4x2+x2  -10  0 0 -4  -6  0 20
    2.  
      graphical ans14
    3.  
      1. y = x3+ 4x2 + x -6
        0 = x3 + 4x2 + x -4
        y = -2
      2. y=0
      3. y = x3 + 4x2 + x - 6
        0 = x3 + 4x2 + 0 – 2
        y = x – 4
        x 1 0 -2
        y -3 -4 -8
        solution 0.8
        -1.5
        And -3.2
        1, -2, -3
  14.  
    1.  
      graphical ans15
    2.  
      1. P = 15.75cm
      2. l=1.5cm
      3. m = 35- 25 = 10 = 6.667
        5.5 – 4.0 1.5
    3. choose P(5,31.4)
      p - 31.4 = 10
      l -5           1.5
      p-31.4 = 100
      l-5          1.5
      15p – 471 = 100k – 500
      15p = 100l – 29
      15           15
      2k = 100
               15
      k= 100 = 3.33
         2 x 15
      c =1.93
  15. P + 0.6 = arh
    Log (P + 0.6) = log a + n log R
    = n log R + log 9
    P + 0.6 1.33 2.65  3.85  8.04  11.22
    Log (P + 0.6) -0.13  0.42  0.59  0.91  1.05
    Log R -0.05  0.05  0.12  0.25  0.30
    Log 0.3 = ¼ = 0.25
    Log a = 0.3
    graphical ans16
  16. x2 + y2 – 6x = 3 – 4y
    x2– 6x + (-6/2)2 + y2 + 4y + (4/2)2 = 3 + (-6/2)2 + (4/2)2
    (x – 3)2 (y + 2)2 = 3 + 9 = 4
    (x – 3)2 (y + 2)2 = 16
    C (3, -2)
    Gradient y = 7 - -2 = 3
                 ∆x     6 – 3
  17.  
    1.  
      x -3  -2  -1  1 2 3 4
      -x3 27  8 1 -1  -8  -27  -64
      2x2  18  8 2 2 8 18 32
      -4x  12  8 4 -4  -8  -12  -16
      2 2 2 2 2 2 2 2
      y 59  26  9 -1  -6  -19  -46
    2. Check on the graph paper.
    3. x = 0.5 + 0.1
    4. –x3 + 2x2 – 5x + 3 = 0
      Line to allow: y = x – 1
      x 0 1
      y -1 0
      x = 0.65
  18. Dy/dx = 12x2 – 12
    12x2 – 12 = 0
    12(x2 – 1) =0
    x = 1
    x = -1
    At x = 1 At x = -1
    0 2 -2  -1  0
    GRD = 12  36  36  0 -12
    -                 0   +    +    0    -
                   (1,7)             (-1, 9)
              Minimum         maximum
    1. table
    2. plotting
      scale
      smooth curve
      1. -0.5 < x < 1 and x>1
      2.  x = 2.5 ±0.1
  19.  2x2 + 2y2 – 6x + 10y + 9 = 0
    x2+ y2 – 3x + 5y + 9/2 = 0
    x2+ y2 – 3x + 5y = -9/2
    x2– 3x + 9/4 + y2 + 5y + 25/4 = 8.5 – 4.5
    (x – 3/2)2 + (y + 5/2)2 = 4
    Radius = 2 units
    Centre = (1.5, -2.5)

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