Gradient intercept form: y = mx + b

Topic Notes
The slope intercept form is used to find the y-intercept and the slope of a straight line. If you look closely, it is a linear function too; meaning the function can tell you its domain and range.

Slope Intercept Form y = mx + b

What is Slope Intercept Form

Slope intercept form is one of the three forms we can use to express a straight line. The other forms are called point slope form and standard form, but we will mostly be using slope intercept form in this section. Using slope intercept form, we express the equation of a line to be:

y=mx+by = mx + b

You may know xx and yy to be coordinates of a point on a graph, but what are mm and bb?

What is b in y=mx+b?

The letter b is a number that represents when the line touches the y-axis. We also refer to this as the "y-intercept". For example, let's draw a straight line on the coordinate plane.

Draw a straight line on the coordinate plane
Draw a straight line on the coordinate plane

If you were to look closely at the y-axis, the straight line touches the y-axis at a specific place. Where is that place? That would be the number 3 because that is where the y-axis and the line intersect. This means we can conclude that b = 3.

What is m in slope intercept form?

The letter m is a number that represents the slope of the line. Some people refer to the slope as rise over run. Recall that if we have two points, then we are able to find the slope of the two points by using the slope formula

m=riserun=y2y1x2x1m = \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

The same idea works here. If we take any two points on a straight line, then we can find the slope of the line using the above formula! For example, let's use this line.

Pick any two points on the straight line to find the slope
Pick any two points on the straight line to find the slope

Notice the points (2, 3) and (0, 1) are on this graph. So why don't we use these two points to find the slope of the line? Using the formula, we would get:

m=y2y1x2x1=1302m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - 3}{0 - 2}

=22= \frac{-2}{-2}

=1= 1

That means the slope of this line is 11!

You can also use the rise over run concept here as well. To go from the point (0, 1) to (2, 3), we have to go 22 units right and 22 units up. That means rise is 22 and run is 22, thus 22=1\frac{2}{2} = 1.

The special thing about slopes is that we can use any two points on the line to find it. So if you took two different points on this line, you would still get that the slope is 11.

How to write an equation in slope intercept form?

You may know what the slope intercept form looks like, but half the time you will be given equations that are not in that form. So it is your job to turn it into slope-intercept form. How do we do that? The goal is to always isolate the yy term. For example, let's say you're given the equation

6x+4+2y=06x + 4 + 2y = 0

To isolate yy, we move the 6x+46x + 4 to the right side of the equation

2y=6x42y = -6x - 4

Now the 22 is in the way of yy, so we are going to get rid of it by dividing both sides of the equation by 22.

2y2=6x42\frac{2y}{2} = \frac{-6x - 4}{2}

y=6x242y = -\frac{6x}{2} - \frac{4}{2}

y=3x2y = -3x - 2

Since yy is isolated, you can see that it is in slope intercept form y=mx+by = mx + b where m=3m = -3, and b=2b = -2.

Now that we know the y intercept and slope really well, why don't we look at specific questions about finding them!

How to find y intercept?

Question 1: Using the linear equation y=12x+5y = \frac{1}{2} x + 5, find y intercept.

Notice here that the equation is already in the slope intercept form y=mx+by = mx + b. We just need to find out what bb is. We can see that b=5b = 5, so the y intercept is 55.

Let's do a slightly harder question.

Question 2: Determine the y-intercept of 2x4y=82x - 4y = 8

Now this linear equation is not in slope intercept form, so we have to change it into that form first. Our goal is to isolate yy in this equation.

See that if we move the 2x2x to the right side of equation, we will have:

4y=82x-4y = 8 - 2x

Now dividing both sides by 4-4, we will get:

4y4=82x4\frac{-4y}{-4} = \frac{8 - 2x}{-4}

y=842x4y = \frac{8}{-4} - \frac{2x}{-4}

y=2+12xy = -2 + \frac{1}{2}x

Now switching the positions of the two terms gives us:

y=12x2y = \frac{1}{2}x - 2

We can clearly see that the equation is in slope intercept form y=mx+by = mx + b. Just by looking at the equation, we can see that =2 = -2, and so the y intercept is 2-2. Let's do another similar question.

Question 3: Determine the y intercept of 4y8=04y - 8 = 0.

This may look a bit weird because there is no xx term, but our goal remains the same. We are going to isolate yy.

Moving the 8-8 to the right side of the equation gives us:

4y=84y = 8

Dividing both sides of the equation by 44 gives us

4y4=84\frac{4y}{4} = \frac{8}{4}

y=2y = 2

Now this may not look like it, but the equation is in slope intercept form. It's just that m=0m = 0, so the entire mxmx term has disappeared. Just rewrite the equation as

y=0m+2y = 0m + 2

From observing, you can tell that b=2b = 2, and so the y intercept is 22. Let's do one more question.

Question 4: Determine (if possible) the y intercept of 5x15=05x - 15 = 0.

This one is interesting because the equation has no yy term. So how are we supposed to put it in slope intercept form? Well, the only thing we can do right now is isolate for xx, so let's try that for now.

Moving the 1515 to the right hand side of equation we have:

5x=155x = 15

Dividing both sides of the equation gives:

5x5=155\frac{5x}{5} = \frac{15}{5}

x=3x = 3

Now we are going to draw this on a graph. Notice that in this equation, xx is forced to be 33 and cannot be anything else. However, it doesn't say anything about yy, so yy can be anything it likes. If we were to write a table of values, we get:

Table of values where x is a contant and y can be anything
Table of values where x is a constant

If we plot these points on a coordinate plane and draw the line, we have:

Plot the points on the coordinate grid where x is a constant and y can be any value
Plot the points on the coordinate grid where x is a constant

Notice how the line never touches the y-axis. This means the equation does not have a y intercept. Now that we covered all the cases of finding the y intercept, let's look at questions which ask us to find the slope!

How to find the slope of an equation?

Question 5: Find the slope of y=32x+1y = \frac{3}{2}x + 1

Notice here that this is in the slope intercept form y=mx+by = mx + b, so by observing, we already know that m=32m = \frac{3}{2}. Hence, the slope is 32\frac{3}{2}!

Question 6: Determine the slope of the linear equation 6x6y=06x - 6y = 0

As you can see, the equation is not in slope intercept form, so we have to convert it into that form first. Our goal is to isolate yy.

Moving 6x6x to the right side of the equation gives:

6y=6x-6y = -6x

Dividing both sides by 6-6 gives:

6y6=6x6\frac{-6y}{-6} = \frac{-6x}{-6}

y=xy = x

Notice that this is actually in slope intercept form y=mx+by = mx + b. It's just that the y intercept bb in this case is 00, and xx is the same as 1x1x. So we can rewrite the equation to:

y=1x+0y = 1x + 0

Now by observing, we see that m=1m = 1. Since mm is the slope, then the slope must be 11. Let's do a slightly harder one

Question 7: Determine the slope of 2y4=02y - 4 = 0

Again, this is a little weird because we have no xx term. However, our goal to isolate yy stays the same.

Moving 4-4 to the right side of the equation we have:

2y=42y = 4

Dividing both sides of the equation by 22 gives:

2y2=42\frac{2y}{2} = \frac{4}{2}

y=2y = 2

See here that it is now in slope intercept form, except the term mxmx is hidden because m=0m = 0. So we can rewrite our equation as:

y=0x+2y = 0x + 2

Since m=0m = 0, then we have a zero slope. If you are wondering what a line with 00 slope looks like, then here is a graph for you to see.

How a graph with zero slope looks like
Graph with a slope of 0

Question 8: Find (if possible) the slope of the linear equation 164x=016 - 4x = 0

In this case, yy cannot be isolated because there is no yy term. So the only thing we can do is to isolate xx.

Moving 1616 to the right side of the equation gives:

4x=16-4x = -16

Dividing both sides by 4-4, we get:

4x4=164\frac{-4x}{-4} = \frac{-16}{-4}

x=4x = 4

This is still not in slope intercept form, so our only hope of attaining the slope is to draw a graph of this line. Again, we see that xx is always forced to be 44, but yy can be anything it likes because there is no yy term. If we were to write a table of values, we get:

Table of values where x is constant and y can be any value
Table of values where x is constant

If we plot these points on a coordinate plane and draw the line, we have:

Plot the points on the coordinate grid where x = 4
Plot the points on the coordinate grid where x = 4

This is a vertical line. So what is the slope of a vertical line? Let's try to figure that out by finding the rise and the run. See how this line is always rising infinitely, but there is no run whatsoever. So that means run is 00. So if we calculate the slope, then we will get:

m=riserun=rise0=undefinedm = \frac{rise}{run} = \frac{rise}{0} = undefined

We cannot divide by 00, so we actually have an undefined slope.

What is an undefined slope?

An undefined slope is a slope that goes straight up in the graph. As seen in the graph above, the slope rises infinitely and has no run. As a result, we get an undefined slope because we cannot divide by 00.

In general, we always get an undefined slope whenever we get a straight vertical line!

Let's look at some other unique questions!

Finding equation from one point

Question 9: A point (2, 6) passes through an equation of y=5x+by = -5x + b. Find "bb".

Notice here that (2, 6) is a coordinate that tells us when x=2x = 2, then y=6y = 6. So to solve for bb, we simply plug the xx and yy values into the equation.

6=5(2)+b6 = -5(2) + b

Isolating and solving for bb gives:

6=10+b6 = -10 + b

b=16b = 16

Recall that bb is also known as the y-intercept, so the y intercept is 1616 as well!

Finding slope from two points

Question 10: Given two points (6, 1) and (-10, 9), find the slope of the line.

Recall that to find the slope of the line, we use the slope equation

m=y2y1x2x1m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Hence using this formula gives us:

m=91106=816m = \frac{9 - 1}{-10 - 6} = \frac{8}{-16}

=12= \frac{1}{-2}

What if we are supposed to find the entire equation of a line instead?

Equation of a line given two points

Question 11: Given two points (-6, 1) and (2, 6), find the slope intercept form equation.

We are basically trying to find the equation in the form of y=mx+by = mx + b. In order to do this, we need to look for mm and bb.

Recall that to find mm, we use the slope equation

m=y2y1x2x1m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Hence using this formula gives us:

m=612(6)=58m = \frac{6 - 1}{2 - (-6)} = \frac{5}{8}

58\frac{5}{8}

So now we have the equation

y=58x+by = \frac{5}{8}x + b

Now we have to look for bb. To solve bb, we pick either of the given points and plug it into the equation. We can do that because both points lie on the line, and any points on the line would satisfy the equation. Let's use the point (2, 6). See that:

6=58(2)+b6 = \frac{5}{8}(2) + b

Isolating bb gives:

6=108+b6 = \frac{10}{8} + b

b=6108b = 6 - \frac{10}{8}

b=488108b = \frac{48}{8} - \frac{10}{8}

b=388b = \frac{38}{8}

Putting this in decimal form, we get that b=4.75b = 4.75. Hence, our slope intercept form equation is:

y=58x+4.75y = \frac{5}{8}x + 4.75

The last thing to cover in this section is to find the domain and range of a line.

How to find domain and range?

To find the domain of a line, we are basically asking ourselves this question: what can xx be? If xx can be those values, then we add them into the domain.

The same thing goes for range. What can yy be? If yy can be those values, then we add them in the range. Let's do an example.

Question 12: Find the domain and range of the equation y=2x+1y = 2x + 1.

Notice that if we draw the graph of this line, then we will get:

Plot the graph of y = 2x + 1
Plot the graph of y = 2x + 1

What can xx be in this line? Notice that xx can be anything because with any xx value, we can get a point that is on the line. The same goes for y. We can always pick a yy value that gives us a point on a line. So we say that

Domain: {x\inR}

Range: {y\inR}

where R means "all real numbers". Let's do a harder one.

Question 13: Find the domain and range of the equation y=2y = -2.

Now if we draw this line on a graph, we will get:

Plot the line of y = -2
Plot the line of y = -2

Notice that xx can be anything because with any xx value, we can get a point that is on the line as long as y=2y = -2. However, look at yy. You see that yy is forced to 2-2 and cannot be anything else. The moment you pick another yy value (like 11), then that point is going to be out of the line. So that means:

Domain: {x \in R}

Range: {y = -2}

Question 14: Find the domain and range of the equation x=1x = 1.

Now if we draw this line on a graph, we will get:

Plot the graph of x = 1
Plot the graph of x = 1

You see that xx is forced to 11 and cannot be anything else. The moment you pick another xx value (like 22), then that point is going to be out of the line.However, look at yy. Notice that yy can be anything because with any yy value, we can get a point that is on the line as long as x=1x = 1

So that means:

Domain: {x=1}

Range: {y\inR}

If you had a lot of problems drawing the graphs to obtain the domain and range, I recommend you use this calculator.

https://www.desmos.com/calculator/2rnqgoa6a4

It teaches you how to graph a linear equation. All you have to do is type the values of mm and bb in. Then it will automatically draw the line for you! This is also useful when you are trying to find the slope intercept form.

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