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 stationary x-coordinate of the quadratic curve y = 3x² − 48x + 3 where dy/dx = 0.
y = 3x² − 48x + 3 · dy/dx = 6x − 48 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 7 where dy/dx = 0.
y = 2x² − 52x + 7 · dy/dx = 4x − 52 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 21 where dy/dx = 0.
y = 6x² − 24x + 21 · dy/dx = 12x − 24 = 0
4. Find the minimum y-value of the quadratic curve y = 5x² − 40x + 23 where dy/dx = 0.
y = 5x² − 40x + 23 · dy/dx = 10x − 40 = 0
5. Find the minimum y-value of the quadratic curve y = 1x² − 24x + 12 where dy/dx = 0.
y = 1x² − 24x + 12 · dy/dx = 2x − 24 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 80x + 14 where dy/dx = 0.
y = 5x² − 80x + 14 · dy/dx = 10x − 80 = 0
7. Find the minimum y-value of the quadratic curve y = 5x² − 90x + 24 where dy/dx = 0.
y = 5x² − 90x + 24 · dy/dx = 10x − 90 = 0
8. Find the minimum y-value of the quadratic curve y = 2x² − 56x + 25 where dy/dx = 0.
y = 2x² − 56x + 25 · dy/dx = 4x − 56 = 0
9. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 10 where dy/dx = 0.
y = 7x² − 56x + 10 · dy/dx = 14x − 56 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 18x + 13 where dy/dx = 0.
y = 1x² − 18x + 13 · dy/dx = 2x − 18 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 2x² − 20x + 6 where dy/dx = 0.
y = 2x² − 20x + 6 · dy/dx = 4x − 20 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 4x² − 24x + 5 where dy/dx = 0.
y = 4x² − 24x + 5 · dy/dx = 8x − 24 = 0

12 questions · Seed 104729 · 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 stationary x-coordinate of the quadratic curve y = 3x² − 48x + 3 where dy/dx = 0.
8
Setting dy/dx = 6x − 48 = 0 gives x = 8.
2. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 7 where dy/dx = 0.
13
Setting dy/dx = 4x − 52 = 0 gives x = 13.
3. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 21 where dy/dx = 0.
2
Setting dy/dx = 12x − 24 = 0 gives x = 2.
4. Find the minimum y-value of the quadratic curve y = 5x² − 40x + 23 where dy/dx = 0.
-57
Setting dy/dx = 10x − 40 = 0 gives x = 4. Substituting x = 4 gives y = 5(4)² − 40(4) + 23 = -57.
5. Find the minimum y-value of the quadratic curve y = 1x² − 24x + 12 where dy/dx = 0.
-132
Setting dy/dx = 2x − 24 = 0 gives x = 12. Substituting x = 12 gives y = 1(12)² − 24(12) + 12 = -132.
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 80x + 14 where dy/dx = 0.
8
Setting dy/dx = 10x − 80 = 0 gives x = 8.
7. Find the minimum y-value of the quadratic curve y = 5x² − 90x + 24 where dy/dx = 0.
-381
Setting dy/dx = 10x − 90 = 0 gives x = 9. Substituting x = 9 gives y = 5(9)² − 90(9) + 24 = -381.
8. Find the minimum y-value of the quadratic curve y = 2x² − 56x + 25 where dy/dx = 0.
-367
Setting dy/dx = 4x − 56 = 0 gives x = 14. Substituting x = 14 gives y = 2(14)² − 56(14) + 25 = -367.
9. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 10 where dy/dx = 0.
-102
Setting dy/dx = 14x − 56 = 0 gives x = 4. Substituting x = 4 gives y = 7(4)² − 56(4) + 10 = -102.
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 18x + 13 where dy/dx = 0.
9
Setting dy/dx = 2x − 18 = 0 gives x = 9.
11. Find the stationary x-coordinate of the quadratic curve y = 2x² − 20x + 6 where dy/dx = 0.
5
Setting dy/dx = 4x − 20 = 0 gives x = 5.
12. Find the stationary x-coordinate of the quadratic curve y = 4x² − 24x + 5 where dy/dx = 0.
3
Setting dy/dx = 8x − 24 = 0 gives x = 3.

12 questions · Seed 104729 · 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 stationary x-coordinate of the quadratic curve y = 3x² − 48x + 3 where dy/dx = 0.
y = 3x² − 48x + 3 · dy/dx = 6x − 48 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 7 where dy/dx = 0.
y = 2x² − 52x + 7 · dy/dx = 4x − 52 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 21 where dy/dx = 0.
y = 6x² − 24x + 21 · dy/dx = 12x − 24 = 0
4. Find the minimum y-value of the quadratic curve y = 5x² − 40x + 23 where dy/dx = 0.
y = 5x² − 40x + 23 · dy/dx = 10x − 40 = 0
5. Find the minimum y-value of the quadratic curve y = 1x² − 24x + 12 where dy/dx = 0.
y = 1x² − 24x + 12 · dy/dx = 2x − 24 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 80x + 14 where dy/dx = 0.
y = 5x² − 80x + 14 · dy/dx = 10x − 80 = 0
7. Find the minimum y-value of the quadratic curve y = 5x² − 90x + 24 where dy/dx = 0.
y = 5x² − 90x + 24 · dy/dx = 10x − 90 = 0
8. Find the minimum y-value of the quadratic curve y = 2x² − 56x + 25 where dy/dx = 0.
y = 2x² − 56x + 25 · dy/dx = 4x − 56 = 0
9. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 10 where dy/dx = 0.
y = 7x² − 56x + 10 · dy/dx = 14x − 56 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 18x + 13 where dy/dx = 0.
y = 1x² − 18x + 13 · dy/dx = 2x − 18 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 2x² − 20x + 6 where dy/dx = 0.
y = 2x² − 20x + 6 · dy/dx = 4x − 20 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 4x² − 24x + 5 where dy/dx = 0.
y = 4x² − 24x + 5 · dy/dx = 8x − 24 = 0

12 questions · Seed 104729 · 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 stationary x-coordinate of the quadratic curve y = 3x² − 48x + 3 where dy/dx = 0.
8
Setting dy/dx = 6x − 48 = 0 gives x = 8.
2. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 7 where dy/dx = 0.
13
Setting dy/dx = 4x − 52 = 0 gives x = 13.
3. Find the stationary x-coordinate of the quadratic curve y = 6x² − 24x + 21 where dy/dx = 0.
2
Setting dy/dx = 12x − 24 = 0 gives x = 2.
4. Find the minimum y-value of the quadratic curve y = 5x² − 40x + 23 where dy/dx = 0.
-57
Setting dy/dx = 10x − 40 = 0 gives x = 4. Substituting x = 4 gives y = 5(4)² − 40(4) + 23 = -57.
5. Find the minimum y-value of the quadratic curve y = 1x² − 24x + 12 where dy/dx = 0.
-132
Setting dy/dx = 2x − 24 = 0 gives x = 12. Substituting x = 12 gives y = 1(12)² − 24(12) + 12 = -132.
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 80x + 14 where dy/dx = 0.
8
Setting dy/dx = 10x − 80 = 0 gives x = 8.
7. Find the minimum y-value of the quadratic curve y = 5x² − 90x + 24 where dy/dx = 0.
-381
Setting dy/dx = 10x − 90 = 0 gives x = 9. Substituting x = 9 gives y = 5(9)² − 90(9) + 24 = -381.
8. Find the minimum y-value of the quadratic curve y = 2x² − 56x + 25 where dy/dx = 0.
-367
Setting dy/dx = 4x − 56 = 0 gives x = 14. Substituting x = 14 gives y = 2(14)² − 56(14) + 25 = -367.
9. Find the minimum y-value of the quadratic curve y = 7x² − 56x + 10 where dy/dx = 0.
-102
Setting dy/dx = 14x − 56 = 0 gives x = 4. Substituting x = 4 gives y = 7(4)² − 56(4) + 10 = -102.
10. Find the stationary x-coordinate of the quadratic curve y = 1x² − 18x + 13 where dy/dx = 0.
9
Setting dy/dx = 2x − 18 = 0 gives x = 9.
11. Find the stationary x-coordinate of the quadratic curve y = 2x² − 20x + 6 where dy/dx = 0.
5
Setting dy/dx = 4x − 20 = 0 gives x = 5.
12. Find the stationary x-coordinate of the quadratic curve y = 4x² − 24x + 5 where dy/dx = 0.
3
Setting dy/dx = 8x − 24 = 0 gives x = 3.

12 questions · Seed 104729 · Age guidance, not curriculum certification