Step by step
The question
Question: find the tangent line to y = 2x² at x = 1.
1 Find the point
The line touches the curve at x = 1.
Its y is f(1): put 1 in.
We get 2.
So the point is (1, 2).
2 Find the slope
The slope of the curve is the derivative.
Use the power rule on each term.
Now we have f′(x) = 4x.
At x = 1: put 1 in.
The slope is m = 4.
Careful!
The slope of a line is one number.
f′(x) changes with x, so we put x = 1 in.
3 Rise = m × run
We call the slope m.
m = rise over run.
So the rise is m times the run.
Like 8/2 is 4, so 8 is 4 times 2.
Take any point (x, y) on the line.
We go from (1, 2) to (x, y).
The rise is y − 2: the change in y.
The run is x − 1: the change in x.
Put them in: rise = m × run becomes y − 2 = 4(x − 1).
This is called point-slope form.
It is made from a point and the slope.
4 Write as y = mx + b
The 4 in front of the bracket multiplies every part inside.
So we write 4 times each part.
Work out each multiplication.
Now we have y − 2 = 4x − 4.
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.
− 2 and + 2 make zero.
Cross them out.
−4 plus 2 is −2.
So y = 4x − 2.
5 Check
Let's check on a picture.
The curve y = 2x².
Our line touches it at (1, 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.