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 = 7x² − 70x + 7 where dy/dx = 0.
y = 7x² − 70x + 7 · dy/dx = 14x − 70 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 1x² − 16x + 14 where dy/dx = 0.
y = 1x² − 16x + 14 · dy/dx = 2x − 16 = 0
3. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 13 where dy/dx = 0.
y = 1x² − 6x + 13 · dy/dx = 2x − 6 = 0
4. Find the minimum y-value of the quadratic curve y = 4x² − 120x + 25 where dy/dx = 0.
y = 4x² − 120x + 25 · dy/dx = 8x − 120 = 0
5. Find the minimum y-value of the quadratic curve y = 4x² − 96x + 4 where dy/dx = 0.
y = 4x² − 96x + 4 · dy/dx = 8x − 96 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 7 where dy/dx = 0.
y = 7x² − 84x + 7 · dy/dx = 14x − 84 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 8x² − 192x + 15 where dy/dx = 0.
y = 8x² − 192x + 15 · dy/dx = 16x − 192 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 4x² − 112x + 16 where dy/dx = 0.
y = 4x² − 112x + 16 · dy/dx = 8x − 112 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 1x² − 26x + 19 where dy/dx = 0.
y = 1x² − 26x + 19 · dy/dx = 2x − 26 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 16 where dy/dx = 0.
y = 5x² − 90x + 16 · dy/dx = 10x − 90 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 8x² − 112x + 9 where dy/dx = 0.
y = 8x² − 112x + 9 · dy/dx = 16x − 112 = 0
12. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 3 where dy/dx = 0.
y = 1x² − 6x + 3 · dy/dx = 2x − 6 = 0

12 questions · Seed 44 · 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 = 7x² − 70x + 7 where dy/dx = 0.
-168
Setting dy/dx = 14x − 70 = 0 gives x = 5. Substituting x = 5 gives y = 7(5)² − 70(5) + 7 = -168.
2. Find the stationary x-coordinate of the quadratic curve y = 1x² − 16x + 14 where dy/dx = 0.
8
Setting dy/dx = 2x − 16 = 0 gives x = 8.
3. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 13 where dy/dx = 0.
4
Setting dy/dx = 2x − 6 = 0 gives x = 3. Substituting x = 3 gives y = 1(3)² − 6(3) + 13 = 4.
4. Find the minimum y-value of the quadratic curve y = 4x² − 120x + 25 where dy/dx = 0.
-875
Setting dy/dx = 8x − 120 = 0 gives x = 15. Substituting x = 15 gives y = 4(15)² − 120(15) + 25 = -875.
5. Find the minimum y-value of the quadratic curve y = 4x² − 96x + 4 where dy/dx = 0.
-572
Setting dy/dx = 8x − 96 = 0 gives x = 12. Substituting x = 12 gives y = 4(12)² − 96(12) + 4 = -572.
6. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 7 where dy/dx = 0.
6
Setting dy/dx = 14x − 84 = 0 gives x = 6.
7. Find the stationary x-coordinate of the quadratic curve y = 8x² − 192x + 15 where dy/dx = 0.
12
Setting dy/dx = 16x − 192 = 0 gives x = 12.
8. Find the stationary x-coordinate of the quadratic curve y = 4x² − 112x + 16 where dy/dx = 0.
14
Setting dy/dx = 8x − 112 = 0 gives x = 14.
9. Find the stationary x-coordinate of the quadratic curve y = 1x² − 26x + 19 where dy/dx = 0.
13
Setting dy/dx = 2x − 26 = 0 gives x = 13.
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 16 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
11. Find the stationary x-coordinate of the quadratic curve y = 8x² − 112x + 9 where dy/dx = 0.
7
Setting dy/dx = 16x − 112 = 0 gives x = 7.
12. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 3 where dy/dx = 0.
-6
Setting dy/dx = 2x − 6 = 0 gives x = 3. Substituting x = 3 gives y = 1(3)² − 6(3) + 3 = -6.

12 questions · Seed 44 · 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 = 7x² − 70x + 7 where dy/dx = 0.
y = 7x² − 70x + 7 · dy/dx = 14x − 70 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 1x² − 16x + 14 where dy/dx = 0.
y = 1x² − 16x + 14 · dy/dx = 2x − 16 = 0
3. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 13 where dy/dx = 0.
y = 1x² − 6x + 13 · dy/dx = 2x − 6 = 0
4. Find the minimum y-value of the quadratic curve y = 4x² − 120x + 25 where dy/dx = 0.
y = 4x² − 120x + 25 · dy/dx = 8x − 120 = 0
5. Find the minimum y-value of the quadratic curve y = 4x² − 96x + 4 where dy/dx = 0.
y = 4x² − 96x + 4 · dy/dx = 8x − 96 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 7 where dy/dx = 0.
y = 7x² − 84x + 7 · dy/dx = 14x − 84 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 8x² − 192x + 15 where dy/dx = 0.
y = 8x² − 192x + 15 · dy/dx = 16x − 192 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 4x² − 112x + 16 where dy/dx = 0.
y = 4x² − 112x + 16 · dy/dx = 8x − 112 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 1x² − 26x + 19 where dy/dx = 0.
y = 1x² − 26x + 19 · dy/dx = 2x − 26 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 16 where dy/dx = 0.
y = 5x² − 90x + 16 · dy/dx = 10x − 90 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 8x² − 112x + 9 where dy/dx = 0.
y = 8x² − 112x + 9 · dy/dx = 16x − 112 = 0
12. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 3 where dy/dx = 0.
y = 1x² − 6x + 3 · dy/dx = 2x − 6 = 0

12 questions · Seed 44 · 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 = 7x² − 70x + 7 where dy/dx = 0.
-168
Setting dy/dx = 14x − 70 = 0 gives x = 5. Substituting x = 5 gives y = 7(5)² − 70(5) + 7 = -168.
2. Find the stationary x-coordinate of the quadratic curve y = 1x² − 16x + 14 where dy/dx = 0.
8
Setting dy/dx = 2x − 16 = 0 gives x = 8.
3. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 13 where dy/dx = 0.
4
Setting dy/dx = 2x − 6 = 0 gives x = 3. Substituting x = 3 gives y = 1(3)² − 6(3) + 13 = 4.
4. Find the minimum y-value of the quadratic curve y = 4x² − 120x + 25 where dy/dx = 0.
-875
Setting dy/dx = 8x − 120 = 0 gives x = 15. Substituting x = 15 gives y = 4(15)² − 120(15) + 25 = -875.
5. Find the minimum y-value of the quadratic curve y = 4x² − 96x + 4 where dy/dx = 0.
-572
Setting dy/dx = 8x − 96 = 0 gives x = 12. Substituting x = 12 gives y = 4(12)² − 96(12) + 4 = -572.
6. Find the stationary x-coordinate of the quadratic curve y = 7x² − 84x + 7 where dy/dx = 0.
6
Setting dy/dx = 14x − 84 = 0 gives x = 6.
7. Find the stationary x-coordinate of the quadratic curve y = 8x² − 192x + 15 where dy/dx = 0.
12
Setting dy/dx = 16x − 192 = 0 gives x = 12.
8. Find the stationary x-coordinate of the quadratic curve y = 4x² − 112x + 16 where dy/dx = 0.
14
Setting dy/dx = 8x − 112 = 0 gives x = 14.
9. Find the stationary x-coordinate of the quadratic curve y = 1x² − 26x + 19 where dy/dx = 0.
13
Setting dy/dx = 2x − 26 = 0 gives x = 13.
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 16 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
11. Find the stationary x-coordinate of the quadratic curve y = 8x² − 112x + 9 where dy/dx = 0.
7
Setting dy/dx = 16x − 112 = 0 gives x = 7.
12. Find the minimum y-value of the quadratic curve y = 1x² − 6x + 3 where dy/dx = 0.
-6
Setting dy/dx = 2x − 6 = 0 gives x = 3. Substituting x = 3 gives y = 1(3)² − 6(3) + 3 = -6.

12 questions · Seed 44 · Age guidance, not curriculum certification