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

Step by step

The question

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

y = 2x² − 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) = 2(2)² − 3(2) = 8 − 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) = 2 × 2x^(2 − 1) − 3 × 1

Now we have f′(x) = 4x − 3.

= 4x − 3

At x = 2: put 2 in.

The slope is m = 5.

f′(2) = 4(2) − 3 = 8 − 3 = 5

Careful!

The slope of a line is one number.

f′(x) changes with x, so we put x = 2 in.

Careful: not m = 4x − 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 = 5(x − 2).

y − 2 = 5(x − 2)

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 5 in front of the bracket multiplies every part inside.

So we write 5 times each part.

y − 2 = 5 × x + 5 × (−2)

Work out each multiplication.

Now we have y − 2 = 5x − 10.

y − 2 = 5x − 10

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.

y ~~− 2 + 2~~ = 5x − 10 + 2

− 2 and + 2 make zero.

Cross them out.

−10 plus 2 is −8.

So y = 5x − 8.

y = 5x − 8

5 Check

Let's check on a picture.

The curve y = 2x² − 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