Worked examples
The question
Question: sketch the graph of y = x³ − 3x.
y = x3 − 3x
1 Where it crosses the axes
Where it crosses the y-axis: put x = 0.
We get y = 0.
x = 0: y = 0
Where it crosses the x-axis: y = 0.
Every term has an x: take it out.
x(x2 − 3) = 0
So x = 0, or x² = 3: x = plus or minus √3, about 1.73.
x = 0 or ±√3 ≈ ±1.73
2 Turning points
The slope: f′(x) = 3x² − 3, by the power rule.
f′(x) = 3x2 − 3
Turning points: where the slope is 0.
f′(x) = 0
3x2 − 3 = 0
Take out the 3.
3(x2 − 1) = 0
x² − 1 is a difference of two squares: x² minus 1 squared.
So it is (x − 1)(x + 1).
3(x + 1)(x − 1) = 0
a² − b² = (a − b)(a + b), b = 1
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = −1 or x = 1.
x = −1 or x = 1
To tell max from min: f″(x) = 6x.
f″(x) = 6x
f″(−1) is negative: a cap, a maximum.
Its y is 2.
f″(−1) = −6 < 0 max
f″(1) is positive: a cup, a minimum.
Its y is −2.
f″(1) = 6 > 0 min
So: a maximum at (−1, 2), and a minimum at (1, −2).
max (−1, 2)
min (1, −2)
3 Inflection points
The bending changes where f″ is 0.
f″(x) = 0
6x = 0
So x = 0.
Its y is 0.
x = 0, y = f(0) = 0
Careful!
Turning points are where f′ is 0.
The inflection point is where f″ is 0.
They are different points.
Careful: a turning point is not an inflection point
4 The two ends
Far out, the x³ term wins.
On the right the graph goes up for ever.
On the left it goes down.
far right: x3 → +∞
far left: x3 → −∞
ends : ↙ left down, right up ↗
5 Draw the graph
Now plot the key points.
Where it crosses the x-axis.
The maximum.
The minimum.
The inflection point.
Join them with a smooth curve: up to the max, down to the min, then up for ever.
6 Check
Let's check one more point: x = 2 gives y = 2.
It sits on our curve.
Right!
x = 2: y = 2
Let's look back.
First: where it crosses the axes.
Then: turning points.
Then: inflection points.
Then: the two ends.
Then: draw the graph.
Last: check.
Answer ends : ↙ left down, right up ↗
▶ Watch it solved
The question
Question: sketch the graph of y = x³ − 3x + 2.
y = x3 − 3x + 2
1 Where it crosses the axes
Where it crosses the y-axis: put x = 0.
We get y = 2.
x = 0: y = 2
Where it crosses the x-axis: try small numbers.
x = 1 gives 0.
f(1) = 0
Factor out x − 1, and solve what is left: x = −2, or 1.
x = −2 or 1
2 Turning points
The slope: f′(x) = 3x² − 3, by the power rule.
f′(x) = 3x2 − 3
Turning points: where the slope is 0.
f′(x) = 0
3x2 − 3 = 0
Take out the 3.
3(x2 − 1) = 0
x² − 1 is a difference of two squares: x² minus 1 squared.
So it is (x − 1)(x + 1).
3(x + 1)(x − 1) = 0
a² − b² = (a − b)(a + b), b = 1
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = −1 or x = 1.
x = −1 or x = 1
To tell max from min: f″(x) = 6x.
f″(x) = 6x
f″(−1) is negative: a cap, a maximum.
Its y is 4.
f″(−1) = −6 < 0 max
f″(1) is positive: a cup, a minimum.
Its y is 0.
f″(1) = 6 > 0 min
So: a maximum at (−1, 4), and a minimum at (1, 0).
max (−1, 4)
min (1, 0)
3 Inflection points
The bending changes where f″ is 0.
f″(x) = 0
6x = 0
So x = 0.
Its y is 2.
x = 0, y = f(0) = 2
Careful!
Turning points are where f′ is 0.
The inflection point is where f″ is 0.
They are different points.
Careful: a turning point is not an inflection point
4 The two ends
Far out, the x³ term wins.
On the right the graph goes up for ever.
On the left it goes down.
far right: x3 → +∞
far left: x3 → −∞
ends : ↙ left down, right up ↗
5 Draw the graph
Now plot the key points.
Where it crosses the x-axis.
The maximum.
The minimum.
The inflection point.
Join them with a smooth curve: up to the max, down to the min, then up for ever.
6 Check
Let's check one more point: x = 2 gives y = 4.
It sits on our curve.
Right!
x = 2: y = 4
Let's look back.
First: where it crosses the axes.
Then: turning points.
Then: inflection points.
Then: the two ends.
Then: draw the graph.
Last: check.
Answer ends : ↙ left down, right up ↗
▶ Watch it solved
The question
Question: sketch the graph of y = x³ − 6x² + 9x.
y = x3 − 6x2 + 9x
1 Where it crosses the axes
Where it crosses the y-axis: put x = 0.
We get y = 0.
x = 0: y = 0
Where it crosses the x-axis: y = 0.
Every term has an x: take it out.
x(x2 − 6x + 9) = 0
Two numbers that multiply to 9 and add to −6: −3 and −3, the same number twice.
x(x − 3)(x − 3) = 0
So x = 0, or 3.
x = 0 or 3
2 Turning points
The slope: f′(x) = 3x² − 12x + 9, by the power rule.
f′(x) = 3x2 − 12x + 9
Turning points: where the slope is 0.
f′(x) = 0
3x2 − 12x + 9 = 0
Take out the 3.
3(x2 − 4x + 3) = 0
We need two numbers that multiply to 3 and add to −4: −1 and −3.
3(x − 1)(x − 3) = 0
−1 × (−3) = 3, −1 + (−3) = −4
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = 1 or x = 3.
x = 1 or x = 3
To tell max from min: f″(x) = 6x − 12.
f″(x) = 6x − 12
f″(1) is negative: a cap, a maximum.
Its y is 4.
f″(1) = −6 < 0 max
f″(3) is positive: a cup, a minimum.
Its y is 0.
f″(3) = 6 > 0 min
So: a maximum at (1, 4), and a minimum at (3, 0).
max (1, 4)
min (3, 0)
3 Inflection points
The bending changes where f″ is 0.
f″(x) = 0
6x − 12 = 0
So x = 2.
Its y is 2.
x = 2, y = f(2) = 2
Careful!
Turning points are where f′ is 0.
The inflection point is where f″ is 0.
They are different points.
Careful: a turning point is not an inflection point
4 The two ends
Far out, the x³ term wins.
On the right the graph goes up for ever.
On the left it goes down.
far right: x3 → +∞
far left: x3 → −∞
ends : ↙ left down, right up ↗
5 Draw the graph
Now plot the key points.
Where it crosses the x-axis.
The maximum.
The minimum.
The inflection point.
Join them with a smooth curve: up to the max, down to the min, then up for ever.
6 Check
Let's check one more point: x = 4 gives y = 4.
It sits on our curve.
Right!
x = 4: y = 4
Let's look back.
First: where it crosses the axes.
Then: turning points.
Then: inflection points.
Then: the two ends.
Then: draw the graph.
Last: check.
Answer ends : ↙ left down, right up ↗
▶ Watch it solved
