Academic Skills Advice
Sketching Curves Summary
When sketching a curve it’s useful to have an idea of the general shape.
𝑦 = 𝑥2
𝑦 = −𝑥 2
𝑦 = 𝑥3 + ⋯
𝑦 = −𝑥 3 + ⋯
Then there are 2 things to think about:
 the 𝒙 and 𝒚 intercepts (where does it cross the axes)
 the turning point(s) (where does it turn and is it max, min or point of inflexion)
The turning point is sometimes called the stationary point or the maximum or minimum.
To sketch a curve
The Turning point(s)
Find the intercepts
To find the y-intercept
To find the position (co-ordinate):
Put 𝒙 = 𝟎
And work out y.
Differentiate once to find
Put y= 𝟎
𝑑𝑥
At a turning point the gradient is always zero.
set
To find the 𝒙-intercept
𝑑𝑦
𝒅𝒚
𝒅𝒙
=𝟎
and solve to find 𝑥.
Substitute into the original equation to find y.
You now have the co-ordinates (𝑥, 𝑦) of any turning
points.
And work out 𝑥.
To find the type (max, min or inflexion):
Differentiate again to find
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
If 𝑑𝑥 2 > 0, it’s a minimum
𝑑2 𝑦
If 𝑑𝑥 2 < 0, it’s a maximum
𝑑2 𝑦
If 𝑑𝑥 2 = 0, normally point of inflexion
but need to check if min or max
© H Jackson 2011 / 2015 / Academic Skills
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Examples:
𝒚 = 𝒙𝟐 − 𝟔𝒙 + 𝟖
Turning point(s)
Intercepts
The y-intercept
Position (co-ordinate):
Put 𝒙 = 𝟎
Differentiate once to find
2
𝑦 = 0 − 6(0) + 8
𝑦=8
set
So it crosses the 𝑦-axis at 8.
𝒅𝒚
𝒅𝒙
𝑑𝑦
𝑑𝑥
= 2𝑥 − 6
=𝟎
2𝑥 − 6 = 0
∴𝑥=𝟑
The 𝒙-intercept
Substitute for 𝑥 in the original to find y:
Put y= 𝟎
𝑦 = 32 − 6(3) + 8
𝑦 = −𝟏
𝑥 2 − 6𝑥 + 8 = 0
Solve to find 𝑥:
You could factorise or use the formula
The turning point is at (3, -1)
(𝑥 − 2)(𝑥 − 4) = 0
𝑥 = 2 𝑜𝑟 𝑥 = 4
Type (max, min or inflexion):
Differentiate again to find
So it crosses the 𝑥-axis at 2 and 4.
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
= 2, positive, so it’s a minimum
Using all of the above information this is what the curve looks like:
𝒚
(Found by putting 𝑥 = 0)
8
(Found by putting y= 0)
𝒙
4
2
(3, -1)
𝑑𝑦
(Found by differentiating and putting 𝑑𝑥 = 0)
© H Jackson 2011 / 2015 / Academic Skills
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𝒚 = 𝒙𝟑 + 𝟐𝒙𝟐 − 𝟑𝒙
Turning point(s)
Intercepts
The y-intercept
Position (co-ordinate):
Put 𝒙 = 𝟎
3
2)
𝑦 = 0 + 2(0
𝑦=0
Differentiate once to find
− 3(0)
set
So it crosses the 𝑦-axis at 0.
𝒅𝒚
𝒅𝒙
𝑑𝑦
𝑑𝑥
= 3𝑥2 + 4𝑥 − 3
=𝟎
3𝑥 2 + 4𝑥 − 3 = 0
Use the quadratic formula to solve for 𝑥:
𝑥 = 𝟎. 𝟒𝟏 𝑜𝑟 𝑥 = −𝟏. 𝟕𝟓
The 𝒙-intercept
Put y= 𝟎
𝑥 3 + 2𝑥 2 − 3𝑥 = 0
Substitute for 𝑥 in the original to find y:
𝑦 = −𝟎. 𝟖𝟑 𝑜𝑟 𝑦 = 𝟔. 𝟎𝟏
Solve to find 𝑥:
The turning points are at
𝑥(𝑥 2 + 2𝑥 − 3) = 0
𝑥(𝑥 + 3)(𝑥 − 1) = 0
𝑥 = 𝟎 𝑜𝑟 𝑥 = −𝟑 𝑜𝑟 𝑥 = 𝟏
Type (max, min or inflexion):
So it crosses the 𝑥-axis at 0, -3 and 1
𝑑2 𝑦
𝑑𝑥 2
(0.41, -0.83) and
(-1.75, 6.01)
= 6𝑥 + 4
𝑥 = 0.41:
𝑥 = −1.75:
6(0.41) + 4 = 6.46 so minimum
6(-1.75) + 4 = -6.5 so maximum
Using all of the above information this is what the curve looks like:
(-1.75, 6.01)
-3
𝒚
0
1
𝒙
(0.41, -0.83)
© H Jackson 2011 / 2015 / Academic Skills
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