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

Step by step

The question

Question: find the tangent line to y = x² − 3x at x = 2.

y = x² − 3x

1 Find the point

The line touches the curve at x = 2.

Its y is f(2): put 2 in.

We get −2.

f(2) = 2² − 3(2) = 4 − 6 = −2

So the point is (2, −2).

point = (2, −2)

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

Now we have f′(x) = 2x − 3.

= 2x − 3

At x = 2: put 2 in.

The slope is m = 1.

f′(2) = 2(2) − 3 = 4 − 3 = 1

Careful!

The slope of a line is one number.

f′(x) changes with x, so we put x = 2 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 (2, −2) to (x, y).

The rise is y − (−2): the change in y.

The run is x − 2: the change in x.

rise = y − (−2)

run = x − 2

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

y − (−2) = (x − 2)

Taking away a negative is the same as adding.

− (−) = +

So it becomes y + 2 = (x − 2).

y + 2 = (x − 2)

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 + 2 = x − 2.

y + 2 = x − 2

We want y by itself.

The + 2 is in the way.

To remove + 2, we take away 2.

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

y ~~+ 2 − 2~~ = x − 2 − 2

+ 2 and − 2 make zero.

Cross them out.

−2 minus 2 is −4.

So y = x − 4.

y = x − 4

5 Check

Let's check on a picture.

The curve y = x² − 3x.

Our line touches it at (2, −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.

Answer