Database - Curve Graph Question for E.Maths

Ahm97sic SG

A Curve Graph Practice Question for E.Maths

A typical "O" level E. Maths curve graph question will consist of 2 parts.

The first part will ask candidates to derive a curve function. The second part will ask the candidates to draw the curve followed by many small parts on the curve graph.

Actual past "O" level E. Maths on Curve Graph

Nov 2000 Paper 2 Question 9

Nov 2001 Paper 2 Question 9

Nov 2002 Paper 2 Question 10

Nov 2003 Paper 2 Question 9

Nov 2004 Paper 2 Question 10

Nov 2005 Paper 2 Question 10

Nov 2006 Paper 2 Question 10

Nov 2007 Paper 2 Question 10

Nov 2008 Paper 2 Question 9 or Question 10 ?

A Curve Graph Practice Question for E.Maths

This practice question encompasses the different types of small parts questions on the curve graph asked in the actual "O" level exams.

Question

The variables x and y are connected by the equation y = x² - x - 5. Some corresponding values of x and y are given in the table below.

x -4 -3 -2 -1 0 1 2 3 4 5

y 15 7 a -3 -5 b -3 1 7 15

(a) Calculate the values of a and b

(b) Use 2 cm to represent 1 unit on the x-axis and 1 cm to represent

1 unit on the y-axis, draw the graph of y = x² - x - 5 for

- 4 £ x £ 5.

(ii) the values of x when y = - 2

(iii) the minimum value of y

(iv) the values of x when y = 0

(v) the value of y when x = 0

(vi) the range of values of y when -2 < x < 2

(d) By drawing a suitable straight line to intersect the curve graph,

find the solutions to the equation

(i) x² - x - 5 = 0

(ii) x² - x - 5 = 2x + 1

(iii) x² - x = 8

(e)* Use your graph to solve the inequality x² - x - 5 > 2x + 1

(f) By drawing a tangent, find the gradient of the curve y = x² - x - 5

at the point (2, -3)

(g)* By drawing a suitable tangent to the curve, find the co-ordinates of

the point A at which the gradient of the tangent at A is 1

(h)* From the graph, find the values of x for which the gradient is positive.

(i) Describe fully the symmetry of the graph.

(j)* Find the cubic equation which is satisfied by the x values of the

intersecting point of the curve x² - x - 5 and the curve 1/x

Hope this will be of help for E. Maths students who will be taking their "O" level exam.

Garrick_3658 SG

(g)* By drawing a suitable tangent to the curve, find the co-ordinates of

the point A at which the gradient of the tangent at A is 1

Do I have to use a set-square? This question makes me wanting to use dy/dx, but fret not, I know I will be penalised :D I'll just it use to double check my answers.

(i) Describe fully the symmetry of the graph.

I had never came across this question in my school papers. Come to think of it, my school has never tested anything on symmetry. Is symmetry still in the E-Maths/A-Maths syllabus?

(j)* Find the cubic equation which is satisfied by the x values of the intersecting point of the curve x² - x - 5 and the curve 1/x

I'll just have to draw the curve y = 1/x and then use the points to find the factors of the cubic equation? Then solve for it by expansion?

Thanks!

Ahm97sic SG

Dear Garrick_3658,

(g)* By drawing a suitable tangent to the curve, find the co-ordinates of

the point A at which the gradient of the tangent at A is 1

Do I have to use a set-square? This question makes me wanting to use dy/dx, but fret not, I know I will be penalised :D I'll just it use to double check my answers.

Yes, I understand that add maths students will be very tempted to use dy/dx to solve this part but it is E.Maths.

To solve it, we need to use a trick ie

Step 1 : Draw a line with gradient of 1 at the origin first.

Gradient of the tangent = 1

= change in y / change in x

= vertical length / horizontal length

= vertcial length of 1 unit / horizontal length of one unit

Start at the origin, one unit horizontal and one unit vertical, we get the point (1, 1)

Join the point (0,0) with (1,1) with a line. This line will have a gradient of one.

Step 2 : Use two rulers or set squares to move the line parallel to just touch the curve

Use two rulers, one perpendicular to the other, and slide until the ruler parallel and just touches the curve and we will get the point (1, -5)

Hence, the co-ordinates of the point A at which the gradient of the tangent at A is 1 is (1, -5).

Regards,

ahm97sic

Ahm97sic SG

Dear Darrick_3658,

(i) Describe fully the symmetry of the graph.

I had never came across this question in my school papers. Come to think of it, my school has never tested anything on symmetry. Is symmetry still in the E-Maths/A-Maths syllabus?

This part of the question combines line and rotational symmetry with the curve graph and its solution. This type of question appeared before in the paper two in June 1992 and June 1993 in "O" level E.Maths.

Regards,

ahm97sic

Ahm97sic SG

Dear Darrick_3658,

(j)* Find the cubic equation which is satisfied by the x values of the intersecting point of the curve x² - x - 5 and the curve 1/x

I'll just have to draw the curve y = 1/x and then use the points to find the factors of the cubic equation? Then solve for it by expansion?

There is a simple trick to find the answer for this question.

Since the question says the cubic equation is satisfied by the x values of the intersecting of the curve x² - x - 5 and the curve 1/x ie we can equate the two curves (due to the intersecting points of x).

Hence, the required cubic equation is

x^2 - x - 5x = 1/x

x^3 - x^2 - 5x = 1

x^3 - x^2 - 5x - 1 = 0

This type of question on curve graph and its solution is common in recent years

"O" level E.Maths.

Regards

ahm97sic

Ahm97sic SG

A Curve Graph Practice Question for E.Maths

A typical "O" level E. Maths curve graph question will consist of 2 parts.

The first part will ask candidates to derive a curve function. The second part will ask the candidates to draw the curve followed by many small parts on the curve graph.

Actual past "O" level E. Maths on Curve Graph

Nov 2000 Paper 2 Question 9

Nov 2001 Paper 2 Question 9

Nov 2002 Paper 2 Question 10

Nov 2003 Paper 2 Question 9

Nov 2004 Paper 2 Question 10

Nov 2005 Paper 2 Question 10

Nov 2006 Paper 2 Question 10

Nov 2007 Paper 2 Question 10

Nov 2008 Paper 2 Question 9 or Question 10 ?

A Curve Graph Practice Question for E.Maths

This practice question encompasses the different types of small parts questions on the curve graph asked in the actual "O" level exams.

Question

The variables x and y are connected by the equation y = x² - x - 5. Some corresponding values of x and y are given in the table below.

x -4 -3 -2 -1 0 1 2 3 4 5

y 15 7 a -3 -5 b -3 1 7 15

(a) Calculate the values of a and b

(b) Use 2 cm to represent 1 unit on the x-axis and 1 cm to represent

1 unit on the y-axis, draw the graph of y = x² - x - 5 for

- 4 £ x £ 5.

(ii) the values of x when y = - 2

(iii) the minimum value of y

(iv) the values of x when y = 0

(v) the value of y when x = 0

(vi) the range of values of y when -2 < x < 2

(d) By drawing a suitable straight line to intersect the curve graph,

find the solutions to the equation

(i) x² - x - 5 = 0

(ii) x² - x - 5 = 2x + 1

(iii) x² - x = 8

(e)* Use your graph to solve the inequality x² - x - 5 > 2x + 1

(f) By drawing a tangent, find the gradient of the curve y = x² - x - 5

at the point (2, -3)

(g)* By drawing a suitable tangent to the curve, find the co-ordinates of

the point A at which the gradient of the tangent at A is 1

(h)* From the graph, find the values of x for which the gradient is

positive.

(i) Describe fully the symmetry of the graph.

(j)* Find the cubic equation which is satisfied by the x values of the

intersecting point of the curve x² - x - 5 and the curve 1/x

Answers

(a) a = 1, b = -5

(b) Please refer to graph

(c)   (i)   y = 10.75
(c)   (ii)   x = 2.30 or -1.30
(c)   (iii) min value of y = -5.25
(c)   (iv) x = 2.79 or -1.79
(c)   (v)   y = -5

(d)   (i)    x = - 1.79 or 2.79
(d)   (ii)   x = - 1.37 or 4.37
(d)   (iii)  x = - 2.37 or 3.37

(e) x > 4.37 or x < 1.37

(f) Gradient = 3

(g) (1, -5)

(h) When x > 0.5, the gradient of the curve is positive.

(i) Curve is symmetrical about the line x = 0.5

(j) x^3 - x^2 - 5x -1 = 0

Thank you for your kind attention.

Regards,

ahm97sic