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.

New variation Try interactive challenge Browse templates

Choose “Save as PDF” to keep a copy. Use your browser’s paper setting for A4 or Letter.

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 + 3 where dy/dx = 0.
y = 2x² − 16x + 3 · dy/dx = 4x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 17 where dy/dx = 0.
y = 7x² − 182x + 17 · dy/dx = 14x − 182 = 0
3. Find the minimum y-value of the quadratic curve y = 2x² − 12x + 13 where dy/dx = 0.
y = 2x² − 12x + 13 · dy/dx = 4x − 12 = 0
4. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 15 where dy/dx = 0.
y = 7x² − 56x + 15 · dy/dx = 14x − 56 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 13 where dy/dx = 0.
y = 5x² − 90x + 13 · dy/dx = 10x − 90 = 0
6. Find the minimum y-value of the quadratic curve y = 7x² − 70x + 13 where dy/dx = 0.
y = 7x² − 70x + 13 · dy/dx = 14x − 70 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 1x² − 30x + 13 where dy/dx = 0.
y = 1x² − 30x + 13 · dy/dx = 2x − 30 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 2x² − 44x + 13 where dy/dx = 0.
y = 2x² − 44x + 13 · dy/dx = 4x − 44 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 7x² − 126x + 13 where dy/dx = 0.
y = 7x² − 126x + 13 · dy/dx = 14x − 126 = 0
10. Find the minimum y-value of the quadratic curve y = 4x² − 104x + 2 where dy/dx = 0.
y = 4x² − 104x + 2 · dy/dx = 8x − 104 = 0
11. Find the minimum y-value of the quadratic curve y = 2x² − 40x + 9 where dy/dx = 0.
y = 2x² − 40x + 9 · dy/dx = 4x − 40 = 0
12. Find the minimum y-value of the quadratic curve y = 5x² − 70x + 9 where dy/dx = 0.
y = 5x² − 70x + 9 · dy/dx = 10x − 70 = 0

12 questions · Seed 43 · 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 + 3 where dy/dx = 0.
-29
Setting dy/dx = 4x − 16 = 0 gives x = 4. Substituting x = 4 gives y = 2(4)² − 16(4) + 3 = -29.
2. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 17 where dy/dx = 0.
13
Setting dy/dx = 14x − 182 = 0 gives x = 13.
3. Find the minimum y-value of the quadratic curve y = 2x² − 12x + 13 where dy/dx = 0.
-5
Setting dy/dx = 4x − 12 = 0 gives x = 3. Substituting x = 3 gives y = 2(3)² − 12(3) + 13 = -5.
4. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 15 where dy/dx = 0.
-97
Setting dy/dx = 14x − 56 = 0 gives x = 4. Substituting x = 4 gives y = 7(4)² − 56(4) + 15 = -97.
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 13 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
6. Find the minimum y-value of the quadratic curve y = 7x² − 70x + 13 where dy/dx = 0.
-162
Setting dy/dx = 14x − 70 = 0 gives x = 5. Substituting x = 5 gives y = 7(5)² − 70(5) + 13 = -162.
7. Find the stationary x-coordinate of the quadratic curve y = 1x² − 30x + 13 where dy/dx = 0.
15
Setting dy/dx = 2x − 30 = 0 gives x = 15.
8. Find the stationary x-coordinate of the quadratic curve y = 2x² − 44x + 13 where dy/dx = 0.
11
Setting dy/dx = 4x − 44 = 0 gives x = 11.
9. Find the stationary x-coordinate of the quadratic curve y = 7x² − 126x + 13 where dy/dx = 0.
9
Setting dy/dx = 14x − 126 = 0 gives x = 9.
10. Find the minimum y-value of the quadratic curve y = 4x² − 104x + 2 where dy/dx = 0.
-674
Setting dy/dx = 8x − 104 = 0 gives x = 13. Substituting x = 13 gives y = 4(13)² − 104(13) + 2 = -674.
11. Find the minimum y-value of the quadratic curve y = 2x² − 40x + 9 where dy/dx = 0.
-191
Setting dy/dx = 4x − 40 = 0 gives x = 10. Substituting x = 10 gives y = 2(10)² − 40(10) + 9 = -191.
12. Find the minimum y-value of the quadratic curve y = 5x² − 70x + 9 where dy/dx = 0.
-236
Setting dy/dx = 10x − 70 = 0 gives x = 7. Substituting x = 7 gives y = 5(7)² − 70(7) + 9 = -236.

12 questions · Seed 43 · 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 + 3 where dy/dx = 0.
y = 2x² − 16x + 3 · dy/dx = 4x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 17 where dy/dx = 0.
y = 7x² − 182x + 17 · dy/dx = 14x − 182 = 0
3. Find the minimum y-value of the quadratic curve y = 2x² − 12x + 13 where dy/dx = 0.
y = 2x² − 12x + 13 · dy/dx = 4x − 12 = 0
4. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 15 where dy/dx = 0.
y = 7x² − 56x + 15 · dy/dx = 14x − 56 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 13 where dy/dx = 0.
y = 5x² − 90x + 13 · dy/dx = 10x − 90 = 0
6. Find the minimum y-value of the quadratic curve y = 7x² − 70x + 13 where dy/dx = 0.
y = 7x² − 70x + 13 · dy/dx = 14x − 70 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 1x² − 30x + 13 where dy/dx = 0.
y = 1x² − 30x + 13 · dy/dx = 2x − 30 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 2x² − 44x + 13 where dy/dx = 0.
y = 2x² − 44x + 13 · dy/dx = 4x − 44 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 7x² − 126x + 13 where dy/dx = 0.
y = 7x² − 126x + 13 · dy/dx = 14x − 126 = 0
10. Find the minimum y-value of the quadratic curve y = 4x² − 104x + 2 where dy/dx = 0.
y = 4x² − 104x + 2 · dy/dx = 8x − 104 = 0
11. Find the minimum y-value of the quadratic curve y = 2x² − 40x + 9 where dy/dx = 0.
y = 2x² − 40x + 9 · dy/dx = 4x − 40 = 0
12. Find the minimum y-value of the quadratic curve y = 5x² − 70x + 9 where dy/dx = 0.
y = 5x² − 70x + 9 · dy/dx = 10x − 70 = 0

12 questions · Seed 43 · 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 + 3 where dy/dx = 0.
-29
Setting dy/dx = 4x − 16 = 0 gives x = 4. Substituting x = 4 gives y = 2(4)² − 16(4) + 3 = -29.
2. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 17 where dy/dx = 0.
13
Setting dy/dx = 14x − 182 = 0 gives x = 13.
3. Find the minimum y-value of the quadratic curve y = 2x² − 12x + 13 where dy/dx = 0.
-5
Setting dy/dx = 4x − 12 = 0 gives x = 3. Substituting x = 3 gives y = 2(3)² − 12(3) + 13 = -5.
4. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 15 where dy/dx = 0.
-97
Setting dy/dx = 14x − 56 = 0 gives x = 4. Substituting x = 4 gives y = 7(4)² − 56(4) + 15 = -97.
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 13 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
6. Find the minimum y-value of the quadratic curve y = 7x² − 70x + 13 where dy/dx = 0.
-162
Setting dy/dx = 14x − 70 = 0 gives x = 5. Substituting x = 5 gives y = 7(5)² − 70(5) + 13 = -162.
7. Find the stationary x-coordinate of the quadratic curve y = 1x² − 30x + 13 where dy/dx = 0.
15
Setting dy/dx = 2x − 30 = 0 gives x = 15.
8. Find the stationary x-coordinate of the quadratic curve y = 2x² − 44x + 13 where dy/dx = 0.
11
Setting dy/dx = 4x − 44 = 0 gives x = 11.
9. Find the stationary x-coordinate of the quadratic curve y = 7x² − 126x + 13 where dy/dx = 0.
9
Setting dy/dx = 14x − 126 = 0 gives x = 9.
10. Find the minimum y-value of the quadratic curve y = 4x² − 104x + 2 where dy/dx = 0.
-674
Setting dy/dx = 8x − 104 = 0 gives x = 13. Substituting x = 13 gives y = 4(13)² − 104(13) + 2 = -674.
11. Find the minimum y-value of the quadratic curve y = 2x² − 40x + 9 where dy/dx = 0.
-191
Setting dy/dx = 4x − 40 = 0 gives x = 10. Substituting x = 10 gives y = 2(10)² − 40(10) + 9 = -191.
12. Find the minimum y-value of the quadratic curve y = 5x² − 70x + 9 where dy/dx = 0.
-236
Setting dy/dx = 10x − 70 = 0 gives x = 7. Substituting x = 7 gives y = 5(7)² − 70(7) + 9 = -236.

12 questions · Seed 43 · Age guidance, not curriculum certification