Tangent line to y = x^2 + 2 at x = 1 Step by Step

Step by step

The question

Question: find the tangent line to y = x² + 2 at x = 1.

y = x² + 2

1 Find the point

The line touches the curve at x = 1.

Its y is f(1): put 1 in.

We get 3.

f(1) = 1² + 2 = 1 + 2 = 3

So the point is (1, 3).

point = (1, 3)

2 Find the slope

The slope of the curve is the derivative.

Use the power rule on each term.

f′(x) = 2x^(2 − 1) + 0

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, 3) to (x, y).

The rise is y − 3: the change in y.

The run is x − 1: the change in x.

rise = y − 3

run = x − 1

Put them in: rise = m × run becomes y − 3 = 2(x − 1).

y − 3 = 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 − 3 = 2 × x + 2 × (−1)

Work out each multiplication.

Now we have y − 3 = 2x − 2.

y − 3 = 2x − 2

We want y by itself.

The − 3 is in the way.

To remove − 3, we add 3.

We do it on both sides, so they stay equal.

y ~~− 3 + 3~~ = 2x − 2 + 3

− 3 and + 3 make zero.

Cross them out.

−2 plus 3 is 1.

So y = 2x + 1.

y = 2x + 1

5 Check

Let's check on a picture.

The curve y = x² + 2.

Our line touches it at (1, 3), 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.

Answer