MathsAges 16-18

Stationary & turning points of curves

Set the first derivative dy/dx = 0 to locate the stationary point or minimum turning value of a quadratic curve. Review suitability for your learners before teaching.

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Worksheet StudioPractice pack · Ages 16-18

Stationary & turning points of curves

Take your time. Show your thinking.

1. Find the minimum y-value of the quadratic curve y = 5x² − 50x + 14 where dy/dx = 0.
y = 5x² − 50x + 14 · dy/dx = 10x − 50 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 120x + 2 where dy/dx = 0.
y = 5x² − 120x + 2 · dy/dx = 10x − 120 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 9 where dy/dx = 0.
y = 7x² − 84x + 9 · dy/dx = 14x − 84 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 8x² − 48x + 9 where dy/dx = 0.
y = 8x² − 48x + 9 · dy/dx = 16x − 48 = 0
5. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 6 where dy/dx = 0.
y = 8x² − 128x + 6 · dy/dx = 16x − 128 = 0
6. Find the minimum y-value of the quadratic curve y = 8x² − 224x + 1 where dy/dx = 0.
y = 8x² − 224x + 1 · dy/dx = 16x − 224 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 4x² − 120x + 7 where dy/dx = 0.
y = 4x² − 120x + 7 · dy/dx = 8x − 120 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 15 where dy/dx = 0.
y = 5x² − 130x + 15 · dy/dx = 10x − 130 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 8x² − 80x + 8 where dy/dx = 0.
y = 8x² − 80x + 8 · dy/dx = 16x − 80 = 0
10. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 16 where dy/dx = 0.
y = 7x² − 140x + 16 · dy/dx = 14x − 140 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 4x² − 32x + 6 where dy/dx = 0.
y = 4x² − 32x + 6 · dy/dx = 8x − 32 = 0
12. Find the minimum y-value of the quadratic curve y = 3x² − 6x + 9 where dy/dx = 0.
y = 3x² − 6x + 9 · dy/dx = 6x − 6 = 0

12 questions · Seed 42 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 16-18

Stationary & turning points of curves

Answers and reasoning. Review before teaching.

1. Find the minimum y-value of the quadratic curve y = 5x² − 50x + 14 where dy/dx = 0.
-111
Setting dy/dx = 10x − 50 = 0 gives x = 5. Substituting x = 5 gives y = 5(5)² − 50(5) + 14 = -111.
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 120x + 2 where dy/dx = 0.
12
Setting dy/dx = 10x − 120 = 0 gives x = 12.
3. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 9 where dy/dx = 0.
6
Setting dy/dx = 14x − 84 = 0 gives x = 6.
4. Find the stationary x-coordinate of the quadratic curve y = 8x² − 48x + 9 where dy/dx = 0.
3
Setting dy/dx = 16x − 48 = 0 gives x = 3.
5. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 6 where dy/dx = 0.
-506
Setting dy/dx = 16x − 128 = 0 gives x = 8. Substituting x = 8 gives y = 8(8)² − 128(8) + 6 = -506.
6. Find the minimum y-value of the quadratic curve y = 8x² − 224x + 1 where dy/dx = 0.
-1567
Setting dy/dx = 16x − 224 = 0 gives x = 14. Substituting x = 14 gives y = 8(14)² − 224(14) + 1 = -1567.
7. Find the stationary x-coordinate of the quadratic curve y = 4x² − 120x + 7 where dy/dx = 0.
15
Setting dy/dx = 8x − 120 = 0 gives x = 15.
8. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 15 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
9. Find the stationary x-coordinate of the quadratic curve y = 8x² − 80x + 8 where dy/dx = 0.
5
Setting dy/dx = 16x − 80 = 0 gives x = 5.
10. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 16 where dy/dx = 0.
-684
Setting dy/dx = 14x − 140 = 0 gives x = 10. Substituting x = 10 gives y = 7(10)² − 140(10) + 16 = -684.
11. Find the stationary x-coordinate of the quadratic curve y = 4x² − 32x + 6 where dy/dx = 0.
4
Setting dy/dx = 8x − 32 = 0 gives x = 4.
12. Find the minimum y-value of the quadratic curve y = 3x² − 6x + 9 where dy/dx = 0.
6
Setting dy/dx = 6x − 6 = 0 gives x = 1. Substituting x = 1 gives y = 3(1)² − 6(1) + 9 = 6.

12 questions · Seed 42 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 16-18

Stationary & turning points of curves

Take your time. Show your thinking.

1. Find the minimum y-value of the quadratic curve y = 5x² − 50x + 14 where dy/dx = 0.
y = 5x² − 50x + 14 · dy/dx = 10x − 50 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 120x + 2 where dy/dx = 0.
y = 5x² − 120x + 2 · dy/dx = 10x − 120 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 9 where dy/dx = 0.
y = 7x² − 84x + 9 · dy/dx = 14x − 84 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 8x² − 48x + 9 where dy/dx = 0.
y = 8x² − 48x + 9 · dy/dx = 16x − 48 = 0
5. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 6 where dy/dx = 0.
y = 8x² − 128x + 6 · dy/dx = 16x − 128 = 0
6. Find the minimum y-value of the quadratic curve y = 8x² − 224x + 1 where dy/dx = 0.
y = 8x² − 224x + 1 · dy/dx = 16x − 224 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 4x² − 120x + 7 where dy/dx = 0.
y = 4x² − 120x + 7 · dy/dx = 8x − 120 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 15 where dy/dx = 0.
y = 5x² − 130x + 15 · dy/dx = 10x − 130 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 8x² − 80x + 8 where dy/dx = 0.
y = 8x² − 80x + 8 · dy/dx = 16x − 80 = 0
10. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 16 where dy/dx = 0.
y = 7x² − 140x + 16 · dy/dx = 14x − 140 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 4x² − 32x + 6 where dy/dx = 0.
y = 4x² − 32x + 6 · dy/dx = 8x − 32 = 0
12. Find the minimum y-value of the quadratic curve y = 3x² − 6x + 9 where dy/dx = 0.
y = 3x² − 6x + 9 · dy/dx = 6x − 6 = 0

12 questions · Seed 42 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 16-18

Stationary & turning points of curves

Answers and reasoning. Review before teaching.

1. Find the minimum y-value of the quadratic curve y = 5x² − 50x + 14 where dy/dx = 0.
-111
Setting dy/dx = 10x − 50 = 0 gives x = 5. Substituting x = 5 gives y = 5(5)² − 50(5) + 14 = -111.
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 120x + 2 where dy/dx = 0.
12
Setting dy/dx = 10x − 120 = 0 gives x = 12.
3. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 9 where dy/dx = 0.
6
Setting dy/dx = 14x − 84 = 0 gives x = 6.
4. Find the stationary x-coordinate of the quadratic curve y = 8x² − 48x + 9 where dy/dx = 0.
3
Setting dy/dx = 16x − 48 = 0 gives x = 3.
5. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 6 where dy/dx = 0.
-506
Setting dy/dx = 16x − 128 = 0 gives x = 8. Substituting x = 8 gives y = 8(8)² − 128(8) + 6 = -506.
6. Find the minimum y-value of the quadratic curve y = 8x² − 224x + 1 where dy/dx = 0.
-1567
Setting dy/dx = 16x − 224 = 0 gives x = 14. Substituting x = 14 gives y = 8(14)² − 224(14) + 1 = -1567.
7. Find the stationary x-coordinate of the quadratic curve y = 4x² − 120x + 7 where dy/dx = 0.
15
Setting dy/dx = 8x − 120 = 0 gives x = 15.
8. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 15 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
9. Find the stationary x-coordinate of the quadratic curve y = 8x² − 80x + 8 where dy/dx = 0.
5
Setting dy/dx = 16x − 80 = 0 gives x = 5.
10. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 16 where dy/dx = 0.
-684
Setting dy/dx = 14x − 140 = 0 gives x = 10. Substituting x = 10 gives y = 7(10)² − 140(10) + 16 = -684.
11. Find the stationary x-coordinate of the quadratic curve y = 4x² − 32x + 6 where dy/dx = 0.
4
Setting dy/dx = 8x − 32 = 0 gives x = 4.
12. Find the minimum y-value of the quadratic curve y = 3x² − 6x + 9 where dy/dx = 0.
6
Setting dy/dx = 6x − 6 = 0 gives x = 1. Substituting x = 1 gives y = 3(1)² − 6(1) + 9 = 6.

12 questions · Seed 42 · Age guidance, not curriculum certification