# Equation of a Line, Tangent Lines

On this page we hope to demonstrate the following:

Equation of a Line, standard form and point-slope

Ex: y = 3x + 4, y - 7 = 3(x - 1)

Finding x-intercept and y-intercept

Finding the equation of tangent lines

Ex: Find the equation of the tangent line of y = x^{3} at x = 2

### Equation of a line

**Standard Form**

When we try to graph polynomials, we quickly find that a polynomial of single degree, where nothing is squared or square-rooted, is always going to be a line. Let's analyze the line y = 3x + 4, whose equation is written in standard form.

The first thing we notice about the graph of this function is that it has a very steep **slope**. Slope is defined as the change in y-value over the change in x-value between two points on a line, but you can easily think of it as the steepness of the line.

It was stated above that this function is stated in standard form: y = 3x + 4. Lines written in standard form are always of the form

y = mx + b,

where m is the **slope**,

b is the **y-intercept** - the y value where the line intersects the y-axis.

The slope and the y-intercept are the only things that we need in order to know what the graph of the line looks like. To begin with, we start by drawing a point at the y-intercept, which in our example is 4, on the y-axis. Then we draw a line that corresponds to our slope, which goes through that point. In our example, the slope was 3. That means that for every unit that we move to the right on the x-axis, we must move 3 points up along the y-axis. The illustration on the right should help.

**Point-Slope form**

In order to write an equation for a line in standard form, we need to know a slope as well as the y-intercept. More often, however, we won't know the y-intercept until we know the equation for the line. However, we can often write the equation for a line in point-slope form very easily. It only requires that we know the slope of the line, and a single point on the line.

Equations of lines in point slope formula look like y - y_{0} = m(x - x_{0}), where m is the slope, y_{0} is the y-value of the point that we know, and x_{0} is the x-value of the point that we know. Given a point (x_{0},y_{0}) and a slope, we can write the equation of a line.

Point-slope form and standard form equations describe the same line - they just show it in a slightly different form. If we look at the graph of y = 3x + 4 above, we may note that (2,10) is a point on the line. then, by our point-slope form, we may also see that an equation for the same line is y - 10 = 3(x - 2). However,

y - 10 = 3(x - 2)

y - 10 = 3x - 6

+10 +10

y = 3x + 4

Notice that they are really the same equation?

**Important:** **The number attached to x is not always the slope**. It only is if we eliminate the coefficient of y. Suppose we had an equation like 3y = 2x + 4. We must first divide the whole equation by 3 so that the coefficient of y is 1. We then find that the slope of that line is actually 2/3.

**Finding the x-intercept and y-intercept**

It is often advantageous, especially for graphing, to figure out where the graph of the line crosses the x-axis and the y-axis. If we have an equation in standard form, it tells us the y-intercept. However, it's not a problem if we don't have an equation in standard form - the intercepts are really easy to find!

Suppose we want to find the y-intercept of the line 2y + 1 = 3x - 2. **The y-intercept is precisely the y-value when the graph crosses the y-axis, which occurs when x = 0. **All we have to do to find the y-intercept, then, is to plug in x = 0, and then solve for y. In our example, this yields

2y + 1 = 3(0) - 2

2y + 1 = -2

**y = -3/2**

so y = -3/2 is the y-intercept. Similarly, **the x-intercept is the x-value when y = 0**, where the graph crosses the x-axis. To find the x-intercept, we plug in y = 0 and solve for x. In our example, this yields

2(0) + 1 = 3x - 2

1 = 3x - 2

**x = 1**

**Exercises:**

Identify the slope and x and y intercepts of the following equations:

1. y = x - 5

2. 2y = 4x

3. y + 4 = 3x - 2

4. A line with a slope of -1 that passes through the point (3,2)

5. A line with a slope of 3/2 that passes through the point (1,0)

### Tangent Line

A **tangent line** **to a curve** is a line that touches a curve, or a graph of a function, at only a single point. For an example, look at the following diagram.

## Example of Tangents
The following demo demonstrates a tangent line to a function. The function is f(x) = x^2. Note how the tangent line touches the graph of the function only at one point.
Dylan Cashman, July 1, 2008, Created with GeoGebra |

It is very easy to find the equation for the tangent line to a curve at a certain point. Remember that to find an equation for a line, all we need is the **slope** at that point, and the coordinates of a single point on that tangent line. The coordinates of the point where the tangent line meets the curve.

**Finding the slope**

To find the slope of a curve at a certain point, something we'll need to know for the tangent line, we will use the derivative. The derivative of a function gives you the slope of that function, since a derivative gives change over time. For example, suppose we had the function f(x) = x^{2}. The derivative of this function is 2x. Note how it gives you a function for a slope - not a single number. This is because the slope of a curve that is not a line is dependent on where you take it. To find the slope of our tangent line at a point, we just plug in the x-value of that point into our function for the slope.

For example, for the function f(x) = x^{2}, we'll try to find the equation of the tangent line at x = 2. To begin with, we need the slope. The derivative of x^{2} is 2x, so to find the slope at x = 2, we plug x = 2 into 2x, to get the slope to be 4. We now have the **slope**. In order to complete the equation for the line, we also need a point the line goes to. We know that the tangent line touches the function f at x = 2, so we can use that point. At x = 2, the function f(x) has the value f(2) = 2^{2} = 4, so we know that the line passes through the point (2,4). Now we have a **point on the line**. We can use the point-slope formula to find the equation of the tangent line: y - 4 = 4(x - 2).

In conclusion, to find the tangent line at a point x_{0}, you need to:

**1. Take the derivative of your function.**

**2. Plug in your x-value, x _{0},**

**into the derivative to get your slope.**

If the derivative ends up just being a constant, like 3, then just use that as your slope.

**3. Use x _{0} to find out what the y-coordinate of that point is.
**Basically, just plug in x

_{0}to get f(x

_{0}) = y

_{0}. Then that point, (x

_{0},y

_{0}), is a point on your tangent line.

**4. Use point-slope formula to write the equation out: y - y _{0} = m(x - x_{0})**

**Exercises:** Find the equation of the tangent lines of the following functions at the indicated points.

1. f(x) = x^{3} + 2x; x = 0

2. f(x) = 4x^{2} - 4x + 1; x = 1

3. f(x) = 3x - 7; x = -14

4. f(x) = x^{5}; x = 5

5. f(x) = x^{5}; x = 1

6. Why is the slope so much larger in exercise 4 than in exercise 5?