PRACTICE WITH SLOPE

In a rush? Need to skip the concepts and jump right to the slope formula?
(Please come back and explore the beautiful ideas at a later time.)

Consider the equation [beautiful math coming... please be patient] $\,y = mx + b\,$.
As we'll see in this section, every equation of this form graphs as a non-vertical line in the coordinate plane.

In this equation, [beautiful math coming... please be patient] $\,m\,$ is the coefficient of the $\,x$-term, and $\,b\,$ is a constant term.
For example, in the equation [beautiful math coming... please be patient] $\,y = 2x + 3\,$, we have $\,m = 2\,$ and $\,b = 3\,$.

The number $\,m\,$ is called the slope of the line, and gives information about the ‘slant’ of the line.
It answers questions like:   “Is it an uphill or downhill line?   How steep is it?”
The purpose of this section is to explain why this is true.

Why does [beautiful math coming... please be patient] $\,y = mx + b\,$ graph as a line?
Why does the number $\,m\,$ give info about the slant of this line?

The key idea (which we'll prove below) is that the equation [beautiful math coming... please be patient] $\,y = mx + b\,$
defines a very special relationship between $\,x\,$ and $\,y\,$:
equal changes in $\,x\,$ (the input) give rise to equal changes in $\,y\,$ (the output).

Indeed, if $\,x\,$ changes by an amount [beautiful math coming... please be patient] $\,\Delta x\,$
(read aloud as ‘delta ex’ or ‘change in ex’),
then $\,y\,$ changes by $\,m\Delta x\,$.
That is, $\,y\,$ changes $\,m\,$ times as fast as $\,x\,$.
This is a bit symbol-intensive, so let's look at a concrete example.

Suppose that [beautiful math coming... please be patient] $\,m\,$ is $\,2\,$. (See the sketch at right.)
Then, $\,y\,$ changes $\,2\,$ times as fast as $\,x\,$ (that is, twice as fast as $\,x\,$).
If (say) $\,x\,$ changes by $\,1\,$, then $\,y\,$ will change by $\,2(1) = 2\,$.
If (say) $\,x\,$ changes by $\,5\,$, then $\,y\,$ will change by $\,2(5) = 10\,$, and so on.

Here's another way to look at it.
Imagine you're ‘standing on’ a point in the coordinate plane.
Let's force [beautiful math coming... please be patient] $\,y\,$ to change twice as fast as $\,x\,$, and see what new points result:

If you take $\,1\,$ step to the right (i.e., let $\,x\,$ change by $\,1\,$),
then you'll have to take $\,2\,$ steps up (i.e., let $\,y\,$ change by $\,2\,$).
Do it again—one step to the right, two steps up.
Do it again—one step to the right, two steps up.
(Repeat, if you want, with different changes in $\,x\,$.)
Hmmm$\,\ldots\,$ what pattern are these points creating? A line!!

If we repeat the previous exercise with $\,m=100\,$,
think about the line that would result:
one step to the right, $\,100\,$ steps up.
Pretty steep line!
So, clearly, the number $\,m\,$ gives information about the steepness or ‘slant’ of the line.

The next part is a bit technical, so perhaps a break is in order first!

PROOF THAT, IN THE EQUATION $\,y=mx+b\,$,
EQUAL CHANGES IN [beautiful math coming... please be patient] $\,x\,$ GIVE RISE TO EQUAL CHANGES IN $\,y\,$

Let [beautiful math coming... please be patient] $\,(x_1,y_1)\,$ be a point on the graph of $\,y = mx + b\,$, so that $\,y_1 = mx_1 + b\,$.
That is, when you substitute $\,x_1\,$ for $\,x\,$ and $\,y_1\,$ for $\,y\,$, the resulting equation is true.

Let [beautiful math coming... please be patient] $\,x_1\,$ change by an amount $\,\Delta x\,$, to get a new $\,x\,$-value:   $\,x_2 = x_1 + \Delta x\,$

Let $\,y_2\,$ represent the $\,y$-value corresponding to $\,x_2\,$.
Then,

$y_2 = mx_2 + b$(find the $\,y$-value corresponding to $\,x_2\,$;   call it $\,y_2\,$)
     [beautiful math coming... please be patient] $= m(x_1 + \Delta x) + b$(substitute $\,x_1 + \Delta x\,$ for $\,x_2\,$)
     [beautiful math coming... please be patient] $= mx_1 + m\Delta x + b$(multiply out; i.e., use the Distributive Law)
     [beautiful math coming... please be patient] $= (mx_1 + b) + m\Delta x$(re-order and re-group)
     [beautiful math coming... please be patient] $= y_1 + m\Delta x$(look back:   $\,mx_1 + b\,$ is precisely $\,y_1\,$)
Thus, $\,y\,$ has changed by $\,m\Delta x\,$.

Make sure you see this!

The starting point is: $($ $x_1$ $,$ $y_1$ $)$
The ending point is: $($ $x_1 + \Delta x$ $,$ $y_1 + m\Delta x$ $)$
So, when [beautiful math coming... please be patient] $\,x\,$ changes by $\,\Delta x\,$, $\,y\,$ changes by $\,m\Delta x\,$.

A FORMULA FOR SLOPE

Let's summarize things from the previous proof.
We started at [beautiful math coming... please be patient] $\,(x_1,y_1)\,$.
We ended at $\,(x_2,y_2)\,$.

The change in $\,x\,$ is:   $\,\Delta x = x_2 - x_1\,$
A ‘change in $\,x\,$’ is informally called a run, since it is in the left/right direction.
Memory device:   Run to the right! Run to the left!

The change in $\,y\,$ is:   $\,\Delta y = y_2 - y_1\,$
A ‘change in $\,y\,$’ is informally called a rise, since it is in the up/down direction.
Memory device:   Rise up!

And, we learned that the change in $\,y\,$ is $\,m\Delta x\,$:   in equation form, [beautiful math coming... please be patient] $\,\Delta y = m\Delta x\,$.
Solving this last equation for the slope, $\,m\,$, gives us an important formula:

SLOPE OF A LINE
Let [beautiful math coming... please be patient] $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$ be two different points on a non-vertical line.
Then, the slope of this line is: [beautiful math coming... please be patient] $$ m = \text{slope} = \frac{y_2-y_1}{x_2-x_1} = \frac{\Delta y}{\Delta x} = \frac{\text{change in } y}{\text{change in } x} = \frac{\text{rise}}{\text{run}} $$
ANALYSIS OF THE SLOPE FORMULA

There are lots of important things you should know about the slope formula:

horizontal line,
zero slope
vertical line,
no slope
UPHILL LINES, POSITIVE SLOPES
gradual uphill,
small positive slope
steep uphill,
large positive slope
DOWNHILL LINES, NEGATIVE SLOPES
gradual downhill,
small negative slope
steep downhill,
large negative slope
EXAMPLES:
Question:
Find the slope of the line through [beautiful math coming... please be patient] $\,(-1,3)\,$ and $\,(2,-5)\,$.
Solution:
[beautiful math coming... please be patient] $\displaystyle\text{slope} = \frac{-5-3}{2-(-1)} = \frac{-8}{3} = -\frac83$

The answers in this exercise are reported as fractions, in simplest form.
Question:
Find the slope of the line through [beautiful math coming... please be patient] $\,(2,5)\,$ and $\,(-7,5)\,$.
Solution:
[beautiful math coming... please be patient] $\displaystyle\text{slope} = \frac{5-5}{-7-2} = \frac{0}{-9} = 0$

Recall that zero, divided by any nonzero number, is zero.
This is a horizontal line.
Question:
Find the slope of the line through [beautiful math coming... please be patient] $\,(2,5)\,$ and $\,(2,-7)\,$.
Solution:
[beautiful math coming... please be patient] $\displaystyle\text{slope} = \frac{-7-5}{2-2} = \frac{-12}{0} = \text{can't continue}$

Division by zero is not allowed.
This line has no slope.
It is a vertical line.
Question:
Horizontal lines have (circle one):    no slope        zero slope
Solution:
zero slope
Question:
Vertical lines have (circle one):    no slope        zero slope
Solution:
no slope
Question:
Suppose a fraction has a zero in the numerator, and a nonzero denominator.
What is the value of the fraction?
Solution:
zero
Question:
Suppose a fraction has a zero in the denominator, and a nonzero numerator.
What is the value of the fraction?
Solution:
it is not defined
Question:
Start at a point $\,(x,y)\,$ on a line.
To get to a new point, move up $\,3\,$ and to the right $\,4\,$.
What is the slope of the line?
Solution:
Rise is $\,3\,$;   run is $\,4\,$.
$\displaystyle\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{3}{4}$
Question:
Start at a point $\,(x,y)\,$ on a line.
To get to a new point, move down $\,2\,$ and to the left $\,6\,$.
What is the slope of the line?
Solution:
Rise is $\,-2\,$;   run is $\,-6\,$.
$\displaystyle\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{-2}{-6} = \frac13$
Question:
Start at a point $\,(x,y)\,$ on a line.
To get to a new point, move straight up $\,5\,$ units.
What is the slope of the line?
Solution:
no slope (vertical line)
Question:
Start at a point $\,(x,y)\,$ on a line.
To get to a new point, move directly to the left $\,5\,$ units.
What is the slope of the line?
Solution:
zero slope (horizontal line)
Question:
Suppose you are walking along a line, moving from left to right.
You are going uphill.
Then, the slope of the line is (circle one):    positive        negative
Solution:
positive
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Graphing Lines

 
 
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
Answers are reported as fractions in simplest form.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
AVAILABLE MASTERED IN PROGRESS

(MAX is 15; there are 15 different problem types)