Tangent line to y = x^3 + 1 at x = -1 Step by Step

Step by step

The question

Question: find the tangent line to y = x³ + 1 at x = −1.

y = x³ + 1

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) = (−1)³ + 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) + 0

Now we have f′(x) = 3x².

= 3x²

At x = −1: put −1 in.

The slope is m = 3.

f′(−1) = 3(−1)² = 3

Careful!

The slope of a line is one number.

f′(x) changes with x, so we put x = −1 in.

Careful: not m = 3x² ✗

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 = 3(x − (−1)).

y − 0 = 3(x − (−1))

Taking away a negative is the same as adding.

− (−) = +

Taking away 0 changes nothing.

− 0: nothing changes

So it becomes y = 3(x + 1).

y = 3(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 3 in front of the bracket multiplies every part inside.

So we write 3 times each part.

y = 3 × x + 3 × 1

Work out each multiplication.

So y = 3x + 3.

y is already by itself.

y = 3x + 3

5 Check

Let's check on a picture.

The curve y = x³ + 1.

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.

Answer