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 = 5x² − 100x + 23 where dy/dx = 0.
y = 5x² − 100x + 23 · dy/dx = 10x − 100 = 0
2. Find the minimum y-value of the quadratic curve y = 4x² − 16x + 2 where dy/dx = 0.
y = 4x² − 16x + 2 · dy/dx = 8x − 16 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 1x² − 4x + 15 where dy/dx = 0.
y = 1x² − 4x + 15 · dy/dx = 2x − 4 = 0
4. Find the minimum y-value of the quadratic curve y = 3x² − 36x + 20 where dy/dx = 0.
y = 3x² − 36x + 20 · dy/dx = 6x − 36 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 6 where dy/dx = 0.
y = 2x² − 52x + 6 · dy/dx = 4x − 52 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 70x + 11 where dy/dx = 0.
y = 5x² − 70x + 11 · dy/dx = 10x − 70 = 0
7. Find the minimum y-value of the quadratic curve y = 4x² − 56x + 11 where dy/dx = 0.
y = 4x² − 56x + 11 · dy/dx = 8x − 56 = 0
8. Find the minimum y-value of the quadratic curve y = 4x² − 40x + 15 where dy/dx = 0.
y = 4x² − 40x + 15 · dy/dx = 8x − 40 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 6x² − 168x + 20 where dy/dx = 0.
y = 6x² − 168x + 20 · dy/dx = 12x − 168 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 7x² − 42x + 4 where dy/dx = 0.
y = 7x² − 42x + 4 · dy/dx = 14x − 42 = 0
11. Find the minimum y-value of the quadratic curve y = 7x² − 98x + 19 where dy/dx = 0.
y = 7x² − 98x + 19 · dy/dx = 14x − 98 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 12x + 22 where dy/dx = 0.
y = 6x² − 12x + 22 · dy/dx = 12x − 12 = 0

12 questions · Seed 48 · 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 = 5x² − 100x + 23 where dy/dx = 0.
10
Setting dy/dx = 10x − 100 = 0 gives x = 10.
2. Find the minimum y-value of the quadratic curve y = 4x² − 16x + 2 where dy/dx = 0.
-14
Setting dy/dx = 8x − 16 = 0 gives x = 2. Substituting x = 2 gives y = 4(2)² − 16(2) + 2 = -14.
3. Find the stationary x-coordinate of the quadratic curve y = 1x² − 4x + 15 where dy/dx = 0.
2
Setting dy/dx = 2x − 4 = 0 gives x = 2.
4. Find the minimum y-value of the quadratic curve y = 3x² − 36x + 20 where dy/dx = 0.
-88
Setting dy/dx = 6x − 36 = 0 gives x = 6. Substituting x = 6 gives y = 3(6)² − 36(6) + 20 = -88.
5. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 6 where dy/dx = 0.
13
Setting dy/dx = 4x − 52 = 0 gives x = 13.
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 70x + 11 where dy/dx = 0.
7
Setting dy/dx = 10x − 70 = 0 gives x = 7.
7. Find the minimum y-value of the quadratic curve y = 4x² − 56x + 11 where dy/dx = 0.
-185
Setting dy/dx = 8x − 56 = 0 gives x = 7. Substituting x = 7 gives y = 4(7)² − 56(7) + 11 = -185.
8. Find the minimum y-value of the quadratic curve y = 4x² − 40x + 15 where dy/dx = 0.
-85
Setting dy/dx = 8x − 40 = 0 gives x = 5. Substituting x = 5 gives y = 4(5)² − 40(5) + 15 = -85.
9. Find the stationary x-coordinate of the quadratic curve y = 6x² − 168x + 20 where dy/dx = 0.
14
Setting dy/dx = 12x − 168 = 0 gives x = 14.
10. Find the stationary x-coordinate of the quadratic curve y = 7x² − 42x + 4 where dy/dx = 0.
3
Setting dy/dx = 14x − 42 = 0 gives x = 3.
11. Find the minimum y-value of the quadratic curve y = 7x² − 98x + 19 where dy/dx = 0.
-324
Setting dy/dx = 14x − 98 = 0 gives x = 7. Substituting x = 7 gives y = 7(7)² − 98(7) + 19 = -324.
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 12x + 22 where dy/dx = 0.
1
Setting dy/dx = 12x − 12 = 0 gives x = 1.

12 questions · Seed 48 · 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 = 5x² − 100x + 23 where dy/dx = 0.
y = 5x² − 100x + 23 · dy/dx = 10x − 100 = 0
2. Find the minimum y-value of the quadratic curve y = 4x² − 16x + 2 where dy/dx = 0.
y = 4x² − 16x + 2 · dy/dx = 8x − 16 = 0
3. Find the stationary x-coordinate of the quadratic curve y = 1x² − 4x + 15 where dy/dx = 0.
y = 1x² − 4x + 15 · dy/dx = 2x − 4 = 0
4. Find the minimum y-value of the quadratic curve y = 3x² − 36x + 20 where dy/dx = 0.
y = 3x² − 36x + 20 · dy/dx = 6x − 36 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 6 where dy/dx = 0.
y = 2x² − 52x + 6 · dy/dx = 4x − 52 = 0
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 70x + 11 where dy/dx = 0.
y = 5x² − 70x + 11 · dy/dx = 10x − 70 = 0
7. Find the minimum y-value of the quadratic curve y = 4x² − 56x + 11 where dy/dx = 0.
y = 4x² − 56x + 11 · dy/dx = 8x − 56 = 0
8. Find the minimum y-value of the quadratic curve y = 4x² − 40x + 15 where dy/dx = 0.
y = 4x² − 40x + 15 · dy/dx = 8x − 40 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 6x² − 168x + 20 where dy/dx = 0.
y = 6x² − 168x + 20 · dy/dx = 12x − 168 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 7x² − 42x + 4 where dy/dx = 0.
y = 7x² − 42x + 4 · dy/dx = 14x − 42 = 0
11. Find the minimum y-value of the quadratic curve y = 7x² − 98x + 19 where dy/dx = 0.
y = 7x² − 98x + 19 · dy/dx = 14x − 98 = 0
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 12x + 22 where dy/dx = 0.
y = 6x² − 12x + 22 · dy/dx = 12x − 12 = 0

12 questions · Seed 48 · 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 = 5x² − 100x + 23 where dy/dx = 0.
10
Setting dy/dx = 10x − 100 = 0 gives x = 10.
2. Find the minimum y-value of the quadratic curve y = 4x² − 16x + 2 where dy/dx = 0.
-14
Setting dy/dx = 8x − 16 = 0 gives x = 2. Substituting x = 2 gives y = 4(2)² − 16(2) + 2 = -14.
3. Find the stationary x-coordinate of the quadratic curve y = 1x² − 4x + 15 where dy/dx = 0.
2
Setting dy/dx = 2x − 4 = 0 gives x = 2.
4. Find the minimum y-value of the quadratic curve y = 3x² − 36x + 20 where dy/dx = 0.
-88
Setting dy/dx = 6x − 36 = 0 gives x = 6. Substituting x = 6 gives y = 3(6)² − 36(6) + 20 = -88.
5. Find the stationary x-coordinate of the quadratic curve y = 2x² − 52x + 6 where dy/dx = 0.
13
Setting dy/dx = 4x − 52 = 0 gives x = 13.
6. Find the stationary x-coordinate of the quadratic curve y = 5x² − 70x + 11 where dy/dx = 0.
7
Setting dy/dx = 10x − 70 = 0 gives x = 7.
7. Find the minimum y-value of the quadratic curve y = 4x² − 56x + 11 where dy/dx = 0.
-185
Setting dy/dx = 8x − 56 = 0 gives x = 7. Substituting x = 7 gives y = 4(7)² − 56(7) + 11 = -185.
8. Find the minimum y-value of the quadratic curve y = 4x² − 40x + 15 where dy/dx = 0.
-85
Setting dy/dx = 8x − 40 = 0 gives x = 5. Substituting x = 5 gives y = 4(5)² − 40(5) + 15 = -85.
9. Find the stationary x-coordinate of the quadratic curve y = 6x² − 168x + 20 where dy/dx = 0.
14
Setting dy/dx = 12x − 168 = 0 gives x = 14.
10. Find the stationary x-coordinate of the quadratic curve y = 7x² − 42x + 4 where dy/dx = 0.
3
Setting dy/dx = 14x − 42 = 0 gives x = 3.
11. Find the minimum y-value of the quadratic curve y = 7x² − 98x + 19 where dy/dx = 0.
-324
Setting dy/dx = 14x − 98 = 0 gives x = 7. Substituting x = 7 gives y = 7(7)² − 98(7) + 19 = -324.
12. Find the stationary x-coordinate of the quadratic curve y = 6x² − 12x + 22 where dy/dx = 0.
1
Setting dy/dx = 12x − 12 = 0 gives x = 1.

12 questions · Seed 48 · Age guidance, not curriculum certification