Tangent line to y = x^2 + 3x - 1 at x = -1 Step by Step

Step by step

The question

Question: find the tangent line to y = x² + 3x − 1 at x = −1.

y = x² + 3x − 1

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) = (−1)² + 3(−1) − 1 = 1 − 3 − 1 = −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) + 3 × 1 + 0

Now we have f′(x) = 2x + 3.

= 2x + 3

At x = −1: put −1 in.

The slope is m = 1.

f′(−1) = 2(−1) + 3 = −2 + 3 = 1

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 ✗

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) = (x − (−1)).

y − (−3) = (x − (−1))

Taking away a negative is the same as adding.

− (−) = +

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.

m = 1, and 1 times a bracket is just the bracket.

y − y₁ = m(x − x₁)

4 Write as y = mx + b

A bracket with a plus or nothing in front can just be removed.

Now we have y + 3 = x + 1.

y + 3 = x + 1

We want y by itself.

The + 3 is in the way.

To remove + 3, we take away 3.

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

y ~~+ 3 − 3~~ = x + 1 − 3

+ 3 and − 3 make zero.

Cross them out.

1 minus 3 is −2.

So y = x − 2.

y = x − 2

5 Check

Let's check on a picture.

The curve y = x² + 3x − 1.

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