Worked examples
The question
Question: find the tangent line to y = x² at x = 1.
y = x2
1 Find the point
The line touches the curve at x = 1.
Its y is f(1): put 1 in.
We get 1.
f(1) = 12 = 1
So the point is (1, 1).
point = (1, 1)
2 Find the slope
The slope of the curve is the derivative.
Use the power rule on each term.
f′(x) = 2x^(2 − 1)
Now we have f′(x) = 2x.
= 2x
At x = 1: put 1 in.
The slope is m = 2.
f′(1) = 2(1) = 2
Careful!
The slope of a line is one number.
f′(x) changes with x, so we put x = 1 in.
Careful: not m = 2x ✗
3 Rise = m × run
We call the slope m.
m = rise over run.
So the rise is m times the run.
rise = m × run
Like 8/2 is 4, so 8 is 4 times 2.
82 = 4 → 8 = 4 × 2
Take any point (x, y) on the line.
We go from (1, 1) to (x, y).
The rise is y − 1: the change in y.
The run is x − 1: the change in x.
rise = y − 1
run = x − 1
Put them in: rise = m × run becomes y − 1 = 2(x − 1).
y − 1 = 2(x − 1)
This is called point-slope form.
It is made from a point and the slope.
y − y1 = m(x − x1)
4 Write as y = mx + b
The 2 in front of the bracket multiplies every part inside.
So we write 2 times each part.
y − 1 = 2 × x + 2 × (−1)
Work out each multiplication.
Now we have y − 1 = 2x − 2.
y − 1 = 2x − 2
We want y by itself.
The − 1 is in the way.
To remove − 1, we add 1.
We do it on both sides, so they stay equal.
y − 1 + 1 = 2x − 2 + 1
− 1 and + 1 make zero.
Cross them out.
−2 plus 1 is −1.
So y = 2x − 1.
y = 2x − 1
5 Check
Let's check on a picture.
The curve y = x².
Our line touches it at (1, 1), with the same steepness.
Right!
Let's look back.
First: find the point.
Then: find the slope.
Then: rise = m × run.
Then: write as y = mx + b.
Last: check.
▶ Watch it solved
The question
Question: find the tangent line to f(x) = x³ − x at x = 1.
f(x) = x3 − x
1 Find the point
The line touches the curve at x = 1.
Its y is f(1): put 1 in.
We get 0.
f(1) = 13 − 1 = 1 − 1 = 0
So the point is (1, 0).
point = (1, 0)
2 Find the slope
The slope of the curve is the derivative.
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 1
Now we have f′(x) = 3x² − 1.
= 3x2 − 1
At x = 1: put 1 in.
The slope is m = 2.
f′(1) = 3(1)2 − 1 = 3 − 1 = 2
Careful!
The slope of a line is one number.
f′(x) changes with x, so we put x = 1 in.
Careful: not m = 3x² − 1 ✗
3 Rise = m × run
We call the slope m.
m = rise over run.
So the rise is m times the run.
rise = m × run
Like 8/2 is 4, so 8 is 4 times 2.
82 = 4 → 8 = 4 × 2
Take any point (x, y) on the line.
We go from (1, 0) to (x, y).
The rise is y − 0: the change in y.
The run is x − 1: the change in x.
rise = y − 0
run = x − 1
Put them in: rise = m × run becomes y − 0 = 2(x − 1).
y − 0 = 2(x − 1)
Taking away 0 changes nothing.
− 0: nothing changes
So it becomes y = 2(x − 1).
y = 2(x − 1)
This is called point-slope form.
It is made from a point and the slope.
y − y1 = m(x − x1)
4 Write as y = mx + b
The 2 in front of the bracket multiplies every part inside.
So we write 2 times each part.
y = 2 × x + 2 × (−1)
Work out each multiplication.
So y = 2x − 2.
y is already by itself.
y = 2x − 2
5 Check
Let's check on a picture.
The curve y = x³ − x.
Our line touches it at (1, 0), with the same steepness.
Right!
Let's look back.
First: find the point.
Then: find the slope.
Then: rise = m × run.
Then: write as y = mx + b.
Last: check.
▶ Watch it solved
The question
Question: find the tangent line to y = x³ − 3x at x = 2.
y = x3 − 3x
1 Find the point
The line touches the curve at x = 2.
Its y is f(2): put 2 in.
We get 2.
f(2) = 23 − 3(2) = 8 − 6 = 2
So the point is (2, 2).
point = (2, 2)
2 Find the slope
The slope of the curve is the derivative.
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 3 × 1
Now we have f′(x) = 3x² − 3.
= 3x2 − 3
At x = 2: put 2 in.
The slope is m = 9.
f′(2) = 3(2)2 − 3 = 12 − 3 = 9
Careful!
The slope of a line is one number.
f′(x) changes with x, so we put x = 2 in.
Careful: not m = 3x² − 3 ✗
3 Rise = m × run
We call the slope m.
m = rise over run.
So the rise is m times the run.
rise = m × run
Like 8/2 is 4, so 8 is 4 times 2.
82 = 4 → 8 = 4 × 2
Take any point (x, y) on the line.
We go from (2, 2) to (x, y).
The rise is y − 2: the change in y.
The run is x − 2: the change in x.
rise = y − 2
run = x − 2
Put them in: rise = m × run becomes y − 2 = 9(x − 2).
y − 2 = 9(x − 2)
This is called point-slope form.
It is made from a point and the slope.
y − y1 = m(x − x1)
4 Write as y = mx + b
The 9 in front of the bracket multiplies every part inside.
So we write 9 times each part.
y − 2 = 9 × x + 9 × (−2)
Work out each multiplication.
Now we have y − 2 = 9x − 18.
y − 2 = 9x − 18
We want y by itself.
The − 2 is in the way.
To remove − 2, we add 2.
We do it on both sides, so they stay equal.
y − 2 + 2 = 9x − 18 + 2
− 2 and + 2 make zero.
Cross them out.
−18 plus 2 is −16.
So y = 9x − 16.
y = 9x − 16
5 Check
Let's check on a picture.
The curve y = x³ − 3x.
Our line touches it at (2, 2), with the same steepness.
Right!
Let's look back.
First: find the point.
Then: find the slope.
Then: rise = m × run.
Then: write as y = mx + b.
Last: check.
▶ Watch it solved
