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 = 1x² − 22x + 12 where dy/dx = 0.
y = 1x² − 22x + 12 · dy/dx = 2x − 22 = 0
2. Find the minimum y-value of the quadratic curve y = 1x² − 22x + 14 where dy/dx = 0.
y = 1x² − 22x + 14 · dy/dx = 2x − 22 = 0
3. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 16 where dy/dx = 0.
y = 8x² − 128x + 16 · dy/dx = 16x − 128 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 7x² − 140x + 12 where dy/dx = 0.
y = 7x² − 140x + 12 · dy/dx = 14x − 140 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 6 where dy/dx = 0.
y = 5x² − 130x + 6 · dy/dx = 10x − 130 = 0
6. Find the minimum y-value of the quadratic curve y = 6x² − 180x + 19 where dy/dx = 0.
y = 6x² − 180x + 19 · dy/dx = 12x − 180 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 28x + 23 where dy/dx = 0.
y = 7x² − 28x + 23 · dy/dx = 14x − 28 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 20 where dy/dx = 0.
y = 1x² − 14x + 20 · dy/dx = 2x − 14 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 8 where dy/dx = 0.
y = 5x² − 90x + 8 · dy/dx = 10x − 90 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 19 where dy/dx = 0.
y = 5x² − 130x + 19 · dy/dx = 10x − 130 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 180x + 18 where dy/dx = 0.
y = 6x² − 180x + 18 · dy/dx = 12x − 180 = 0
12. Find the minimum y-value of the quadratic curve y = 4x² − 80x + 20 where dy/dx = 0.
y = 4x² − 80x + 20 · dy/dx = 8x − 80 = 0

12 questions · Seed 45 · 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 = 1x² − 22x + 12 where dy/dx = 0.
11
Setting dy/dx = 2x − 22 = 0 gives x = 11.
2. Find the minimum y-value of the quadratic curve y = 1x² − 22x + 14 where dy/dx = 0.
-107
Setting dy/dx = 2x − 22 = 0 gives x = 11. Substituting x = 11 gives y = 1(11)² − 22(11) + 14 = -107.
3. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 16 where dy/dx = 0.
-496
Setting dy/dx = 16x − 128 = 0 gives x = 8. Substituting x = 8 gives y = 8(8)² − 128(8) + 16 = -496.
4. Find the stationary x-coordinate of the quadratic curve y = 7x² − 140x + 12 where dy/dx = 0.
10
Setting dy/dx = 14x − 140 = 0 gives x = 10.
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 6 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
6. Find the minimum y-value of the quadratic curve y = 6x² − 180x + 19 where dy/dx = 0.
-1331
Setting dy/dx = 12x − 180 = 0 gives x = 15. Substituting x = 15 gives y = 6(15)² − 180(15) + 19 = -1331.
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 28x + 23 where dy/dx = 0.
2
Setting dy/dx = 14x − 28 = 0 gives x = 2.
8. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 20 where dy/dx = 0.
7
Setting dy/dx = 2x − 14 = 0 gives x = 7.
9. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 8 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 19 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 180x + 18 where dy/dx = 0.
15
Setting dy/dx = 12x − 180 = 0 gives x = 15.
12. Find the minimum y-value of the quadratic curve y = 4x² − 80x + 20 where dy/dx = 0.
-380
Setting dy/dx = 8x − 80 = 0 gives x = 10. Substituting x = 10 gives y = 4(10)² − 80(10) + 20 = -380.

12 questions · Seed 45 · 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 = 1x² − 22x + 12 where dy/dx = 0.
y = 1x² − 22x + 12 · dy/dx = 2x − 22 = 0
2. Find the minimum y-value of the quadratic curve y = 1x² − 22x + 14 where dy/dx = 0.
y = 1x² − 22x + 14 · dy/dx = 2x − 22 = 0
3. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 16 where dy/dx = 0.
y = 8x² − 128x + 16 · dy/dx = 16x − 128 = 0
4. Find the stationary x-coordinate of the quadratic curve y = 7x² − 140x + 12 where dy/dx = 0.
y = 7x² − 140x + 12 · dy/dx = 14x − 140 = 0
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 6 where dy/dx = 0.
y = 5x² − 130x + 6 · dy/dx = 10x − 130 = 0
6. Find the minimum y-value of the quadratic curve y = 6x² − 180x + 19 where dy/dx = 0.
y = 6x² − 180x + 19 · dy/dx = 12x − 180 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 28x + 23 where dy/dx = 0.
y = 7x² − 28x + 23 · dy/dx = 14x − 28 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 20 where dy/dx = 0.
y = 1x² − 14x + 20 · dy/dx = 2x − 14 = 0
9. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 8 where dy/dx = 0.
y = 5x² − 90x + 8 · dy/dx = 10x − 90 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 19 where dy/dx = 0.
y = 5x² − 130x + 19 · dy/dx = 10x − 130 = 0
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 180x + 18 where dy/dx = 0.
y = 6x² − 180x + 18 · dy/dx = 12x − 180 = 0
12. Find the minimum y-value of the quadratic curve y = 4x² − 80x + 20 where dy/dx = 0.
y = 4x² − 80x + 20 · dy/dx = 8x − 80 = 0

12 questions · Seed 45 · 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 = 1x² − 22x + 12 where dy/dx = 0.
11
Setting dy/dx = 2x − 22 = 0 gives x = 11.
2. Find the minimum y-value of the quadratic curve y = 1x² − 22x + 14 where dy/dx = 0.
-107
Setting dy/dx = 2x − 22 = 0 gives x = 11. Substituting x = 11 gives y = 1(11)² − 22(11) + 14 = -107.
3. Find the minimum y-value of the quadratic curve y = 8x² − 128x + 16 where dy/dx = 0.
-496
Setting dy/dx = 16x − 128 = 0 gives x = 8. Substituting x = 8 gives y = 8(8)² − 128(8) + 16 = -496.
4. Find the stationary x-coordinate of the quadratic curve y = 7x² − 140x + 12 where dy/dx = 0.
10
Setting dy/dx = 14x − 140 = 0 gives x = 10.
5. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 6 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
6. Find the minimum y-value of the quadratic curve y = 6x² − 180x + 19 where dy/dx = 0.
-1331
Setting dy/dx = 12x − 180 = 0 gives x = 15. Substituting x = 15 gives y = 6(15)² − 180(15) + 19 = -1331.
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 28x + 23 where dy/dx = 0.
2
Setting dy/dx = 14x − 28 = 0 gives x = 2.
8. Find the stationary x-coordinate of the quadratic curve y = 1x² − 14x + 20 where dy/dx = 0.
7
Setting dy/dx = 2x − 14 = 0 gives x = 7.
9. Find the stationary x-coordinate of the quadratic curve y = 5x² − 90x + 8 where dy/dx = 0.
9
Setting dy/dx = 10x − 90 = 0 gives x = 9.
10. Find the stationary x-coordinate of the quadratic curve y = 5x² − 130x + 19 where dy/dx = 0.
13
Setting dy/dx = 10x − 130 = 0 gives x = 13.
11. Find the stationary x-coordinate of the quadratic curve y = 6x² − 180x + 18 where dy/dx = 0.
15
Setting dy/dx = 12x − 180 = 0 gives x = 15.
12. Find the minimum y-value of the quadratic curve y = 4x² − 80x + 20 where dy/dx = 0.
-380
Setting dy/dx = 8x − 80 = 0 gives x = 10. Substituting x = 10 gives y = 4(10)² − 80(10) + 20 = -380.

12 questions · Seed 45 · Age guidance, not curriculum certification