Questions

 Use a convenient scale to draw the graph of y = x^{2} + 5x – 3 for the range 2 ≤ x ≤ 6
 Use your graph to determine the roots of the equation 5x – x^{2} – 3 = 0
 Use your graph to solve the equation 2x – x^{2} + 3 = 0 by drawing a suitable straight line
 Find a quadratic equation whose roots are 2.5 + √3 and 2.5 – √3, expressing it in the form ax^{2} + bx + c = 0 Where a, b and c are integers.

 Complete the table below for the equation y = x^{2} + 3x – 6 for 6 ≤ x ≤ 4
x 6 5 4 3 2 1 0 1 2 3 4 y 12 6 6 22  Using a scale 1cm to represent 2 units in both axes. Draw the graph of y = x^{2} + 3x – 6
 Use your graph to solve:
 X^{2} + 3X = 6
 X^{2} + 3X – 2 = 0
 Complete the table below for the equation y = x^{2} + 3x – 6 for 6 ≤ x ≤ 4

 Complete the table for the function: y = 2x^{2} + 3x + 1
x 4 3 2 1 0 1 2 3 2x^{2} 18 0 18 3x+1 7 0 10 y 10 1 6  Use the table in (a) above to draw the graph : y = 2x^{2} + 3x + 1 for 4 ≤ x ≤ 3
 Use the graph in (b) to solve the equation :
 2x^{2} + 4x – 3 = 0
 x^{2} + ^{3}/_{2}x + 2 = 3
 Complete the table for the function: y = 2x^{2} + 3x + 1
 A youth group decided to raise Ksh 480,000 to buy a piece of land costing Ksh. 80,000 per hectare. Before the actual payment was made, four of the members pulled out and each of those remaining had to pay an additional Kshs. 20,000.
 If the original number of the group members was x, write down;
 An expression of how much each was to contribute originally.
 An expression of how the remaining members were to contribute after the four pulled out.
 Determine the number of members who actually contributed towards the purchase of the land.
 Calculate the ratio of the supposed original contribution to the new contribution.
 If the land was subdivided equally, find the size of land each member got. (2 mk)
 If the original number of the group members was x, write down;

 Draw the graph of y = 2x^{2} + x – 2 given the range 3 ≤ x ≤ 2
 Use your graph above to solve
 2x^{2} + x – 2 =0
 2x^{2} + x – 3 =0
 2x^{2} + x5 =0

 Use trapezoidal rule to find the area between the curve y = x^{2} + 4x + 4, the x axis and the coordinates x = 2 and x = 1. Take values of x at intervals of ½ unit.
 Use integration to find the exact area. Hence find the percentage error in your approximation.

 Use trapezoidal rule to find the area between the curve y = x^{2} + 4x + 4, the x axis and the coordinates x = 2 and x = 1. Take values of x at intervals of ½ unit.
 Use integration to find the exact area. Hence find the percentage error in your approximation.
 Draw the graph of y = 2x^{2} – 4x  5 for x between 3 and 5 on the grid provided.
 State the line of symmetry for the graph
 State the range of values for which 2x^{2} – 4x – 5 ≤ 0
 On the same set of axes, draw the graph of y=2x +3
 Determine the solutions to the equation: 2x^{2} – 4x – 5 = 2x +3
 Complete the table below for the equation y = 5 + 3x 2x^{2} by filling in the blank spaces.
X 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5 3 3.5 Y 9 3 6 6 4  Use the values from the table above to draw the graph of y = 5 + 3x 2x^{2} (3mks)
 Use the graph to:
 Find the maximum point of the function 5 + 3x 2x^{2}
 Determine the range of values and give the integral values which satisfy
the inequality 5 + 3x 2x^{2} ≥ 2

 Complete the table below for the function y = 2x^{2} + 4x – 3
x 4 3 2 1 0 1 2 2x^{2} 32 8 2 0 4x3 11 3 5 y 3 3 13  Draw the graph of the function y = 2x^{2} + 4x – 3 and use your graph to estimate the roots of the equation 2x^{2} + 4x – 3 = 0.
 In order to solve graphically the equation 2x^{2} + x – 5 = 0, a straight line must be drawn to intersect the curve y = 2x^{2} + 4x – 3. Determine the equation of this line, draw it and hence obtain the roots of the equation 2x^{2} + x – 5 = 0 to 1 decimal place.
 Complete the table below for the function y = 2x^{2} + 4x – 3

 Complete the table for the function y = 1 – 2x – 3x^{2} for  3 ≤ x ≤ 3.
x 3 2 1 0 1 2 3 3x^{2} 27 3 0 12 2x 4 0 6 1 1 1 1 1 1 1 1 y 20 1 15  Using the table above, draw the graph of y = 1 – 2x – 3x^{2} (Scale 1 cm represent 0.5 units on
xaxis and 1 cm rep 2 units on the y – axis on the grid provided.  Use the graph in (b) above to solve.
 1 – 2x – 3x^{2} = 0
 2 – 5x – 3x^{2} = 0
 Complete the table for the function y = 1 – 2x – 3x^{2} for  3 ≤ x ≤ 3.
 A quadratic equation x^{2} + ax – b = 0 has roots 1 and 5 , determine the values of a and b
 Find a quadratic equation whose roots are 1.5 + √2 and 1.5 – √2 , expressing it in the form ax^{2} + bx + c = 0, where a, b, and c are integers
 If a^{2} + b^{2} = 89 and a + b = 13
 Find the values of;
 a^{2} + 2ab + b^{2}
 2ab
 a^{2} – 2ab + b^{2}
 a  b
 Determine the values of a and b
 Find the values of;
 Complete the table below for the function y = 2x^{3 }+5x^{2 } x  6 (2 mks)
x 4 3 2 1 0 1 2 2x^{3} 128 54 0 2 16 5x^{2} 80 45 20 5 0 5 20 x 4 3 0 1 6 6 6 6 6 6 6 6 y 50 6 0  On the grid provided draw the graph y = 2x^{3 }+5x^{2 } x  6 for 4 ≤x ≤2. Use 2cm to represent 1 unit on the xaxis and 1 cm to represent 5 units on the y – axis (4 mks)
 By drawing a suitable line, use the graph in (b) to solve the
 2x^{3} + 5x^{2} + x  4 = 0
 2x^{3 }+ 5x^{2 } x + 2 = 0
 Complete the table below for the function y = 2x^{3 }+5x^{2 } x  6 (2 mks)
Answers


x 2 1 0 1 2 3 4 5 6 y 17 9 3 1 3 3 1 3 9  y = 5x – x^{2}  3
0 = 5x –x^{2} 3
y = 0
x = 0.75 or 4.3 ± 0.1  y = 5x – x^{2}  3
0 = 2x – x^{2} + 3
y = 3x – 6
x 0 1 2
y 6 9 0
x = 1 or 3 + 0.1

 x – 2.5  √3 x – 2.5 + √3 = 0
x^{2} – 2.5x + x√3  2.5x + 6.25 – 2.5 √3
x√3 + 2.5 √3 = 0
x^{2} – 5x + 6.25 – 3 = 0 

x 6 5 4 3 2 1 0 1 2 3 4 y 04 2 8 8 2 4 12 
 x^{2} + 3x – 6 = 0
x=  4.5 or 1.5 ± 0.2  y = x^{2} + 3x 6
x^{2} + 3x 2
y = 4
 x^{2} + 3x – 6 = 0



x 4 3 2 1 0 1 2 3 y 21 10 3 0 1 6 15 28 

 2x^{2} + 3x + 1 = 0
2x^{2} + 4x – 3 = 0
x + 2 = y
x = 0.6 or x = 2.6 ± 0.1  x = 0.30 –x = 1.8 ± 0.1
 2x^{2} + 3x + 1 = 0



 ^{480,000/=}/_{x}
 ^{48000}/_{(x4)}
 480,000 = 480,000 + 20,000
x – 4 x
Multiply all hr’ by L.C.M.
480,000x = 480,000(x – 4) + 20,000(x^{2} – 4x)
Dividing by 10,000
48x = 48x – 192 + 2x^{2} – 4x
48x – 48x + 4x – 2x^{2} + 192 = 0
4x – 2x^{2} + 192 = 0
But x cannot be –ve hence
x = ^{43.3954}/_{4} = 10.8489
= 11  Original : new cont.
480,000 : 480, 000
11 7  Size of land bought = 6 hectares
^{6}/_{7} = 0.857143
≃ 0.8571 hectares






x 4 3 2 1 0 1 2 2x^{2} 32 18 8 2 0 2 8 4x3 19 15 11 7 3 1 5 y 13 3 3 5 3 3 13  Roots for x = 2.6 ± 0.1
x = 0.6 ± 0.1
y = 2x^{2} + 4x – 3
0 = 2x^{2} + x – 5
y = 3x + 2
Roots read from the 2 pts of intersection of the line and curve.
X =  1.9 + 0.1
X = 1.4 + 0.1


x 3 2 1 0 1 2 3 3x^{2} 27 12 3 0 3 12 27 2x 6 4 2 0 2 4 6 1 1 1 1 1 1 1 1 y 20 7 0 1 4 15 32
1 – 2x – 3x^{2} = 0
x = 1
or x = 0.7
y = 3x^{2} – 2x + 1
0 = 3x^{2} – 5x + 2
y = 0 + 3x – 1
x 0 2
y 1 5
 x^{2} + ax – b = 0
(x1) (x +5) = x^{2} + ax – b
x^{2} + 4x – 5 = x^{2} + ax –b
a = 4, b = 5  Let a = 1.5 + √2
b = 1.5  √2
(x – a) (x – b) = 0
x^{2} – xb – ax + ab = 0
x^{2} – x (1.5  √2) – x (1.5 + √2) + ab = 0
x^{2} –1.5x +x√2) – x 1.5x √2) = 0
x^{2} –3x + ab
x^{2} –3x + (1.5 + √2) (1.5  √2) = 0
x^{2} –3x + 2.25  2 = 0
x^{2} –3x + ¼ = 0
4x^{2} – 12x + 1 = 0 

 a^{2} + b^{2} = 89 a + b = 13
a^{2} + 2ab + b^{2} = (a + b)^{2} = 13^{2} = 169  2ab = 169 – 89
= 80  a^{2} – 2ab + b^{2} = a^{2} + b^{2} – 2ab
= 89 – 80 = 9  (a – b)^{2} = 9
ab = ±3
 a^{2} + b^{2} = 89 a + b = 13
 a + b = 13
a – b = 3
2a = 16

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