Finding Tangent Lines Examples

Find the tangent line to f(x) = x² at x = 1.

We need a point and a slope.

Since f(1) = 1, the point we want is (1, 1). Since we found earlier that f ' (1) = 2, 2 is the slope we want.

We want a line of the form

y = mx + b whose slope is 2, so m = 2: y = 2x + b.

We also want (1, 1) to be on the line, so

1 = 2(1) + b.

Solving, we find b = -1. This means the equation for the tangent line to f at 1 is

y = 2x – 1.

To check this answer, we graph the function f(x) = x² and the line y = 2x - 1 on the same graph:

Since the line bounces off the curve at x = 1, this looks like a reasonable answer.

When finding equations for tangent lines, check the answers. Stick both the original function and the tangent line in the calculator, and make sure the picture looks right. If we find something like this, we know we've made a mistake somewhere:

If we find something that looks like a tangent line and quacks like a tangent line, there's a good chance we've correctly found the tangent line.

Since we haven't discussed the shortcuts for finding derivatives yet, these exercises will require derivatives we've already found. 

We can find the derivatives again for practice or we can go look them up.

Find the equation for the tangent line to f at a.

There are no numbers, but don't panic. This will work the same way as all the other problems. We need a point and a slope. Since we don't have a formula, we can't figure out what f(a) is, but we know the point (a, f(a)) is the one we want. Since we don't have a formula, we can't figure out what
f ' (a) is, but we know f ' (a) is the slope we want. We want a line of the form

y = mx + b

whose slope is f ' (a), so m = f ' (a):

y = f ' (a)x + b.

We also want (a, f(a)) to be on the line, so

f(a) = f ' (a)(a) + b.

Solving, we find b = f(a) – f ' (a)a. This means the equation for the tangent line to f at a is
y = f ' (a)x + (f(a) – f ' (a)a). This is a little messy. We can make it prettier by rearranging. First, remove the unnecessary parentheses:

y = f ' (a)x + f(a) – f ' (a)a

Then swap the second and third terms:

y = f ' (a)x – f ' (a)a + f(a)

Pull out f ' (a) from the first two terms:

And there we are. The magic formula says the equation for the tangent line is

y = f ' (a)(x – a) + f(a).

We might also see it written

y = f(a) + f ' (a)(x – a),

which is the same thing with the terms switched.

Since we've given in and explained the magic formula, we should probably show how to use it, too.

Here, we aren't as nice. We'll need to find the derivatives from scratch, using the sneaky methods.

Let f(x) = 3 + x + x². Use the magic formula to find the tangent line to f at a = 1.

We know a = 1, so f(a) = f(1) = 3 + 1 + 1 = 5. Now we need to find f ' (1). With a = 1,

Since f(1) = 5, we have

Now we have all the pieces: a = 1, f(a) = 5, and f ' (a) = 3. Once everything's in the formula

y = f(a) + f ' (a)(x – a)

to find y = 5 + 3(x – 1).

We may or may not want to "simplify" this to find y = 3x + 2.

The picture below shows a function and its tangent line at x = 2: 

What is f ' (2)?

f ' (2) is the slope of the tangent line to f at 2. Since we have two points on the tangent line, we can find its slope:

This means f ' (2) = 0.5.