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 = 2x² − 16x + 7 where dy/dx = 0.
y = 2x² − 16x + 7 · dy/dx = 4x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 5 where dy/dx = 0.
y = 5x² − 130x + 5 · dy/dx = 10x − 130 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 5x² − 150x + 24 where dy/dx = 0.
y = 5x² − 150x + 24 · dy/dx = 10x − 150 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 2x² − 8x + 25 where dy/dx = 0.
y = 2x² − 8x + 25 · dy/dx = 4x − 8 = 0
5. Find the minimum y-value of the quadratic curve y = 7x² − 84x + 2 where dy/dx = 0.
y = 7x² − 84x + 2 · dy/dx = 14x − 84 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 22 where dy/dx = 0.
y = 1x² − 14x + 22 · dy/dx = 2x − 14 = 0
7. Find the minimum y-value of the quadratic curve y = 4x² − 88x + 12 where dy/dx = 0.
y = 4x² − 88x + 12 · dy/dx = 8x − 88 = 0
8. Find the minimum y-value of the quadratic curve y = 5x² − 110x + 2 where dy/dx = 0.
y = 5x² − 110x + 2 · dy/dx = 10x − 110 = 0
9. Find the minimum y-value of the quadratic curve y = 3x² − 78x + 10 where dy/dx = 0.
y = 3x² − 78x + 10 · dy/dx = 6x − 78 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 10x + 19 where dy/dx = 0.
y = 1x² − 10x + 19 · dy/dx = 2x − 10 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 72x + 15 where dy/dx = 0.
y = 6x² − 72x + 15 · dy/dx = 12x − 72 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 12 where dy/dx = 0.
y = 6x² − 24x + 12 · dy/dx = 12x − 24 = 0

12 questions · Seed 46 · 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 = 2x² − 16x + 7 where dy/dx = 0.
-25
Setting dy/dx = 4x − 16 = 0 gives x = 4. Substituting x = 4 gives y = 2(4)² − 16(4) + 7 = -25.
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 5 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
3. Find the stationary x-coordinate of the quadratic curve y = 5x² − 150x + 24 where dy/dx = 0.
15
Setting dy/dx = 10x − 150 = 0 gives x = 15.
4. Find the stationary x-coordinate of the quadratic curve y = 2x² − 8x + 25 where dy/dx = 0.
2
Setting dy/dx = 4x − 8 = 0 gives x = 2.
5. Find the minimum y-value of the quadratic curve y = 7x² − 84x + 2 where dy/dx = 0.
-250
Setting dy/dx = 14x − 84 = 0 gives x = 6. Substituting x = 6 gives y = 7(6)² − 84(6) + 2 = -250.
6. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 22 where dy/dx = 0.
7
Setting dy/dx = 2x − 14 = 0 gives x = 7.
7. Find the minimum y-value of the quadratic curve y = 4x² − 88x + 12 where dy/dx = 0.
-472
Setting dy/dx = 8x − 88 = 0 gives x = 11. Substituting x = 11 gives y = 4(11)² − 88(11) + 12 = -472.
8. Find the minimum y-value of the quadratic curve y = 5x² − 110x + 2 where dy/dx = 0.
-603
Setting dy/dx = 10x − 110 = 0 gives x = 11. Substituting x = 11 gives y = 5(11)² − 110(11) + 2 = -603.
9. Find the minimum y-value of the quadratic curve y = 3x² − 78x + 10 where dy/dx = 0.
-497
Setting dy/dx = 6x − 78 = 0 gives x = 13. Substituting x = 13 gives y = 3(13)² − 78(13) + 10 = -497.
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 10x + 19 where dy/dx = 0.
5
Setting dy/dx = 2x − 10 = 0 gives x = 5.
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 72x + 15 where dy/dx = 0.
6
Setting dy/dx = 12x − 72 = 0 gives x = 6.
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 12 where dy/dx = 0.
2
Setting dy/dx = 12x − 24 = 0 gives x = 2.

12 questions · Seed 46 · 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 = 2x² − 16x + 7 where dy/dx = 0.
y = 2x² − 16x + 7 · dy/dx = 4x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 5 where dy/dx = 0.
y = 5x² − 130x + 5 · dy/dx = 10x − 130 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 5x² − 150x + 24 where dy/dx = 0.
y = 5x² − 150x + 24 · dy/dx = 10x − 150 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 2x² − 8x + 25 where dy/dx = 0.
y = 2x² − 8x + 25 · dy/dx = 4x − 8 = 0
5. Find the minimum y-value of the quadratic curve y = 7x² − 84x + 2 where dy/dx = 0.
y = 7x² − 84x + 2 · dy/dx = 14x − 84 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 22 where dy/dx = 0.
y = 1x² − 14x + 22 · dy/dx = 2x − 14 = 0
7. Find the minimum y-value of the quadratic curve y = 4x² − 88x + 12 where dy/dx = 0.
y = 4x² − 88x + 12 · dy/dx = 8x − 88 = 0
8. Find the minimum y-value of the quadratic curve y = 5x² − 110x + 2 where dy/dx = 0.
y = 5x² − 110x + 2 · dy/dx = 10x − 110 = 0
9. Find the minimum y-value of the quadratic curve y = 3x² − 78x + 10 where dy/dx = 0.
y = 3x² − 78x + 10 · dy/dx = 6x − 78 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 10x + 19 where dy/dx = 0.
y = 1x² − 10x + 19 · dy/dx = 2x − 10 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 72x + 15 where dy/dx = 0.
y = 6x² − 72x + 15 · dy/dx = 12x − 72 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 12 where dy/dx = 0.
y = 6x² − 24x + 12 · dy/dx = 12x − 24 = 0

12 questions · Seed 46 · 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 = 2x² − 16x + 7 where dy/dx = 0.
-25
Setting dy/dx = 4x − 16 = 0 gives x = 4. Substituting x = 4 gives y = 2(4)² − 16(4) + 7 = -25.
2. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 5 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
3. Find the stationary x-coordinate of the quadratic curve y = 5x² − 150x + 24 where dy/dx = 0.
15
Setting dy/dx = 10x − 150 = 0 gives x = 15.
4. Find the stationary x-coordinate of the quadratic curve y = 2x² − 8x + 25 where dy/dx = 0.
2
Setting dy/dx = 4x − 8 = 0 gives x = 2.
5. Find the minimum y-value of the quadratic curve y = 7x² − 84x + 2 where dy/dx = 0.
-250
Setting dy/dx = 14x − 84 = 0 gives x = 6. Substituting x = 6 gives y = 7(6)² − 84(6) + 2 = -250.
6. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 22 where dy/dx = 0.
7
Setting dy/dx = 2x − 14 = 0 gives x = 7.
7. Find the minimum y-value of the quadratic curve y = 4x² − 88x + 12 where dy/dx = 0.
-472
Setting dy/dx = 8x − 88 = 0 gives x = 11. Substituting x = 11 gives y = 4(11)² − 88(11) + 12 = -472.
8. Find the minimum y-value of the quadratic curve y = 5x² − 110x + 2 where dy/dx = 0.
-603
Setting dy/dx = 10x − 110 = 0 gives x = 11. Substituting x = 11 gives y = 5(11)² − 110(11) + 2 = -603.
9. Find the minimum y-value of the quadratic curve y = 3x² − 78x + 10 where dy/dx = 0.
-497
Setting dy/dx = 6x − 78 = 0 gives x = 13. Substituting x = 13 gives y = 3(13)² − 78(13) + 10 = -497.
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 10x + 19 where dy/dx = 0.
5
Setting dy/dx = 2x − 10 = 0 gives x = 5.
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 72x + 15 where dy/dx = 0.
6
Setting dy/dx = 12x − 72 = 0 gives x = 6.
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 12 where dy/dx = 0.
2
Setting dy/dx = 12x − 24 = 0 gives x = 2.

12 questions · Seed 46 · Age guidance, not curriculum certification