Step by step
The question
Question: find the tangent line to y = x² at x = 3.
1 Find the point
The line touches the curve at x = 3.
Its y is f(3): put 3 in.
We get 9.
So the point is (3, 9).
2 Find the slope
The slope of the curve is the derivative.
Use the power rule on each term.
Now we have f′(x) = 2x.
At x = 3: put 3 in.
The slope is m = 6.
Careful!
The slope of a line is one number.
f′(x) changes with x, so we put x = 3 in.
3 Use point-slope form
Point-slope form: the line through the point (x₁, y₁) with slope m is y − y₁ = m(x − x₁).
Put in the point (3, 9) and m = 6: y − 9 = 6(x − 3).
4 Write as y = mx + b
We also write it as y = mx + b.
Then we can read the slope and where the line crosses the y-axis.
The 6 in front of the bracket multiplies every part inside.
So we write 6 times each part.
Work out each multiplication.
6 times 3 is 18.
Plus times minus is minus.
Now we have y − 9 = 6x − 18.
Is y alone now?
Not yet: − 9 is still with y.
How do we remove it?
We want y by itself.
The − 9 is in the way.
To remove − 9, we add 9.
We do it on both sides, so they stay equal.
− 9 and + 9 cancel each other.
They make zero.
−18 plus 9 is −9.
So y = 6x − 9.
5 Check
Let's check on a picture.
The curve y = x².
Our line touches it at (3, 9), with the same steepness.
Right!
Let's look back.
First: find the point.
Then: find the slope.
Then: use point-slope form.
Then: write as y = mx + b.
Last: check.