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 = 4x² − 16x + 3 where dy/dx = 0.
y = 4x² − 16x + 3 · dy/dx = 8x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 8x² − 240x + 10 where dy/dx = 0.
y = 8x² − 240x + 10 · dy/dx = 16x − 240 = 0
3. Find the minimum y-value of the quadratic curve y = 3x² − 60x + 7 where dy/dx = 0.
y = 3x² − 60x + 7 · dy/dx = 6x − 60 = 0
4. Find the minimum y-value of the quadratic curve y = 7x² − 210x + 6 where dy/dx = 0.
y = 7x² − 210x + 6 · dy/dx = 14x − 210 = 0
5. Find the minimum y-value of the quadratic curve y = 8x² − 64x + 10 where dy/dx = 0.
y = 8x² − 64x + 10 · dy/dx = 16x − 64 = 0
6. Find the minimum y-value of the quadratic curve y = 1x² − 28x + 9 where dy/dx = 0.
y = 1x² − 28x + 9 · dy/dx = 2x − 28 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 22 where dy/dx = 0.
y = 7x² − 182x + 22 · dy/dx = 14x − 182 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 8x² − 32x + 9 where dy/dx = 0.
y = 8x² − 32x + 9 · dy/dx = 16x − 32 = 0
9. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 18 where dy/dx = 0.
y = 7x² − 140x + 18 · dy/dx = 14x − 140 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 2x² − 48x + 4 where dy/dx = 0.
y = 2x² − 48x + 4 · dy/dx = 4x − 48 = 0
11. 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. Find the stationary x-coordinate of the quadratic curve y = 3x² − 54x + 6 where dy/dx = 0.
y = 3x² − 54x + 6 · dy/dx = 6x − 54 = 0

12 questions · Seed 47 · 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 = 4x² − 16x + 3 where dy/dx = 0.
-13
Setting dy/dx = 8x − 16 = 0 gives x = 2. Substituting x = 2 gives y = 4(2)² − 16(2) + 3 = -13.
2. Find the stationary x-coordinate of the quadratic curve y = 8x² − 240x + 10 where dy/dx = 0.
15
Setting dy/dx = 16x − 240 = 0 gives x = 15.
3. Find the minimum y-value of the quadratic curve y = 3x² − 60x + 7 where dy/dx = 0.
-293
Setting dy/dx = 6x − 60 = 0 gives x = 10. Substituting x = 10 gives y = 3(10)² − 60(10) + 7 = -293.
4. Find the minimum y-value of the quadratic curve y = 7x² − 210x + 6 where dy/dx = 0.
-1569
Setting dy/dx = 14x − 210 = 0 gives x = 15. Substituting x = 15 gives y = 7(15)² − 210(15) + 6 = -1569.
5. Find the minimum y-value of the quadratic curve y = 8x² − 64x + 10 where dy/dx = 0.
-118
Setting dy/dx = 16x − 64 = 0 gives x = 4. Substituting x = 4 gives y = 8(4)² − 64(4) + 10 = -118.
6. Find the minimum y-value of the quadratic curve y = 1x² − 28x + 9 where dy/dx = 0.
-187
Setting dy/dx = 2x − 28 = 0 gives x = 14. Substituting x = 14 gives y = 1(14)² − 28(14) + 9 = -187.
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 22 where dy/dx = 0.
13
Setting dy/dx = 14x − 182 = 0 gives x = 13.
8. Find the stationary x-coordinate of the quadratic curve y = 8x² − 32x + 9 where dy/dx = 0.
2
Setting dy/dx = 16x − 32 = 0 gives x = 2.
9. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 18 where dy/dx = 0.
-682
Setting dy/dx = 14x − 140 = 0 gives x = 10. Substituting x = 10 gives y = 7(10)² − 140(10) + 18 = -682.
10. Find the stationary x-coordinate of the quadratic curve y = 2x² − 48x + 4 where dy/dx = 0.
12
Setting dy/dx = 4x − 48 = 0 gives x = 12.
11. 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. Find the stationary x-coordinate of the quadratic curve y = 3x² − 54x + 6 where dy/dx = 0.
9
Setting dy/dx = 6x − 54 = 0 gives x = 9.

12 questions · Seed 47 · 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 = 4x² − 16x + 3 where dy/dx = 0.
y = 4x² − 16x + 3 · dy/dx = 8x − 16 = 0
2. Find the stationary x-coordinate of the quadratic curve y = 8x² − 240x + 10 where dy/dx = 0.
y = 8x² − 240x + 10 · dy/dx = 16x − 240 = 0
3. Find the minimum y-value of the quadratic curve y = 3x² − 60x + 7 where dy/dx = 0.
y = 3x² − 60x + 7 · dy/dx = 6x − 60 = 0
4. Find the minimum y-value of the quadratic curve y = 7x² − 210x + 6 where dy/dx = 0.
y = 7x² − 210x + 6 · dy/dx = 14x − 210 = 0
5. Find the minimum y-value of the quadratic curve y = 8x² − 64x + 10 where dy/dx = 0.
y = 8x² − 64x + 10 · dy/dx = 16x − 64 = 0
6. Find the minimum y-value of the quadratic curve y = 1x² − 28x + 9 where dy/dx = 0.
y = 1x² − 28x + 9 · dy/dx = 2x − 28 = 0
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 22 where dy/dx = 0.
y = 7x² − 182x + 22 · dy/dx = 14x − 182 = 0
8. Find the stationary x-coordinate of the quadratic curve y = 8x² − 32x + 9 where dy/dx = 0.
y = 8x² − 32x + 9 · dy/dx = 16x − 32 = 0
9. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 18 where dy/dx = 0.
y = 7x² − 140x + 18 · dy/dx = 14x − 140 = 0
10. Find the stationary x-coordinate of the quadratic curve y = 2x² − 48x + 4 where dy/dx = 0.
y = 2x² − 48x + 4 · dy/dx = 4x − 48 = 0
11. 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. Find the stationary x-coordinate of the quadratic curve y = 3x² − 54x + 6 where dy/dx = 0.
y = 3x² − 54x + 6 · dy/dx = 6x − 54 = 0

12 questions · Seed 47 · 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 = 4x² − 16x + 3 where dy/dx = 0.
-13
Setting dy/dx = 8x − 16 = 0 gives x = 2. Substituting x = 2 gives y = 4(2)² − 16(2) + 3 = -13.
2. Find the stationary x-coordinate of the quadratic curve y = 8x² − 240x + 10 where dy/dx = 0.
15
Setting dy/dx = 16x − 240 = 0 gives x = 15.
3. Find the minimum y-value of the quadratic curve y = 3x² − 60x + 7 where dy/dx = 0.
-293
Setting dy/dx = 6x − 60 = 0 gives x = 10. Substituting x = 10 gives y = 3(10)² − 60(10) + 7 = -293.
4. Find the minimum y-value of the quadratic curve y = 7x² − 210x + 6 where dy/dx = 0.
-1569
Setting dy/dx = 14x − 210 = 0 gives x = 15. Substituting x = 15 gives y = 7(15)² − 210(15) + 6 = -1569.
5. Find the minimum y-value of the quadratic curve y = 8x² − 64x + 10 where dy/dx = 0.
-118
Setting dy/dx = 16x − 64 = 0 gives x = 4. Substituting x = 4 gives y = 8(4)² − 64(4) + 10 = -118.
6. Find the minimum y-value of the quadratic curve y = 1x² − 28x + 9 where dy/dx = 0.
-187
Setting dy/dx = 2x − 28 = 0 gives x = 14. Substituting x = 14 gives y = 1(14)² − 28(14) + 9 = -187.
7. Find the stationary x-coordinate of the quadratic curve y = 7x² − 182x + 22 where dy/dx = 0.
13
Setting dy/dx = 14x − 182 = 0 gives x = 13.
8. Find the stationary x-coordinate of the quadratic curve y = 8x² − 32x + 9 where dy/dx = 0.
2
Setting dy/dx = 16x − 32 = 0 gives x = 2.
9. Find the minimum y-value of the quadratic curve y = 7x² − 140x + 18 where dy/dx = 0.
-682
Setting dy/dx = 14x − 140 = 0 gives x = 10. Substituting x = 10 gives y = 7(10)² − 140(10) + 18 = -682.
10. Find the stationary x-coordinate of the quadratic curve y = 2x² − 48x + 4 where dy/dx = 0.
12
Setting dy/dx = 4x − 48 = 0 gives x = 12.
11. 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. Find the stationary x-coordinate of the quadratic curve y = 3x² − 54x + 6 where dy/dx = 0.
9
Setting dy/dx = 6x − 54 = 0 gives x = 9.

12 questions · Seed 47 · Age guidance, not curriculum certification