When working with exponents, there are two new words that come up: exponent and base. As an example, in x^2, 2 is the exponent while x is the base and in  4^{12}, 12 is the exponent while 4 is the base.

exponent-and-base

As long as the exponent is a positive whole number, you can think of it as telling you how many times you should multiply the base by itself.

  2^3=2\cdot2\cdot2=8
  x^3=x\cdot x\cdot x
  5^2=5\cdot5=25

As you can see,   4^{12} means “4 multiplied by itself 12 times” and that is a really big number. Many values that you calculate from exponents will be quite large.

There are a couple of things to note:

  • Anything with an exponent of zero is defined to be 1. So,   4^{0}=1,   100^{0}=1.
  • Ok right above, when I said anything, I meant “almost anything”. 0^0 is an indeterminate form. For everything else however, the rule above holds.
  • An exponent of 1 is the same as just writing the number by itself:   4^{1}=4.

Negative Exponents

Negative exponents are treated a bit differently. By definition, if we have a positive number n and a nonzero b, then the following rule holds:


b^{-n} = \dfrac{1}{b^n}

Using this rule, if you see a value with a negative exponent, then it can be rewritten as 1 “over” that term with a positive exponent. Here are a few examples of how this works.

3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}
10^{-1} = \dfrac{1}{10^1} = \dfrac{1}{10}
(-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16}

As you can see, it is simply about following the formula when dealing with negative exponents. After this, you can simply calculate the value using the rules above.

Fractional Exponents

In algebra and more advanced math, it is very common to see an exponent of one-half or some other fraction. This is one way to represent terms involving roots, or radical terms. By definition (if n is nonzero):


b^{\frac{m}{n}}= \sqrt[n]{b^m}

Working with these types of terms (radical terms) is a more advanced topic and will be the subject of a different article, but it is important that you be able to rewrite such terms using the definition as below.

3^{\frac{1}{2}}=\sqrt{3^1}=\sqrt{3}
4^{\frac{2}{3}}=\sqrt[3]{4^2} = \sqrt[3]{16}

As you study terms involving exponents, you will find that you need to combine rules quite often. For example, the rule above can be combined with the rule for negative exponents to simplify even more complex expressions.

Using a Calculator

Finding the value of terms with exponents on scientific or graphing calculators is quite easy. The usual key you will need will use a carat symbol ^ to represent that the next number is an exponent. So, 3^4 can be represented as 3^4 in most calculators. On the TI83 or TI84:

exponent-on-calculator

screenshot-exponent-calculator

You can also always use the carat symbol on google (just type it into google! they now do automatic calculations for you) or wolfram alpha.

It doesn’t matter what function you are looking at, the y intercept is the point at which the graph of that function crosses the y-axis. That means the point can always be written (0, c) for some number c.


The y-intercept is at y = -1.

The y-intercept is at y = -1. As a point, we would write (0, -1)



Finding y Intercepts

Along the y-axis, the x value of any point is zero. This means that to find the y intercept of any function, we can just let x = 0 and solve for y. For example, consider the function y=\frac{1}{3}x+4. If we let x = 0, then y=\frac{1}{3}(0)+4 = 4. As a point, this would be the point (0, 4).

Similarly, the function in the graph at the top of the page is, y=x^2-1. When x = 0, y = 0^2 - 1 = -1., meaning the y intercept is -1, the same as on the graph. If this is still confusing, in the video to the right, I go through a couple more examples.

A Closer Look at the y Intercept

Now that we have seen how to find them, there are two interesting questions that can come up:

  1. Can a function have more than one y intercept?
  2. Can a function have no y intercept?

In answering these, remember that a function can only have one output (y-value) for each input (x-value). If a function has more than one y intercept, that means that there are two y’s for the input x = 0. This isn’t possible, so no, it is not possible for a function to have more than one y-intercept.

What about no y intercept? Well, take a look at the graph of y=\frac{1}{x} below. It never crosses the y-axis, so it has no y-intercept.


This function has no y-intercept.

This function has no y-intercept.


What happened? If you let x=0, you get y = \frac{1}{0} which is undefined. This is just one example, but there are other similar functions, so yes, it is certainly possible for a function to not have a y-intercept.

Using a Graphing Calculator

Graphing calculators can be used to find x and y intercepts pretty quickly. A while back, I made a short video that can show you how this works. You can find it on the mathbootcamps youtube page.

What About x-Intercepts?

Well, you can take a look at my discussion on x intercepts right over here!

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Given the graph of any function like y = x – 2, the x intercept is simply the point where the graph crosses the x-axis. There might be just one such point or many. In either case, the idea stays the same. Another way to think about this is, the x intercept is the point on the graph where y = 0. This can be used to easily find the x intercept for any graph. In this guide, we will talk about how to find x intercepts using a graph, an equation, and a graphing calculator.

Finding the x-intercept or intercepts using a Graph

As mentioned above, functions may have one, zero, or even many x intercepts. These can be found by looking at where the graph of a function crosses the x-axis. This point (or these points) is graphed for each of the functions below.


Graph of y = x - 2
The x intercept is 2

The x intercept is 2



Graph of y = x^2 + 1
The x intercepts are -1 and 1.

The x intercepts are -1 and 1.



Graph of y = x^4-8x^3-49x^2+260x+300
The x intercepts are -6, -1, 5, and 10.

The x intercepts are -6, -1, 5, and 10.


How to Find x Intercepts

When the graph of a function is crossing the x-axis, what does the point look like as an ordered pair? For instance, is the x-intercept is said to be 5, what point is this? Well, since there is no “height” to that point, we would write it as (5, 0). In fact, any x-intercept would have a 0 as the y-value.

We can use this to find the x-intercept of intercepts of any function if we have the equation. The general rule is let y = 0 and solve for x. How easy this is depends entirely on the equation!

As an example, let’s use the first equation graphed above:  y = x - 2. If we let y = 0, then we get the equation 0 = x - 2 This equation can be solved by using the usual rules for solving linear equations: we just add 2 to both sides and find the x intercept is x = 2.

Things are a little more complicated with the second example. If  y = 0 in the equation y = x^2 - 1, then we get the quadratic equation 0 = x^2 - 1. Adding 1 to both sides, we find that x^2 = 1 which has solutions 1 and -1 by the square root rule.

Of course, the idea remains the same even when we get to the last (much more complicated) equation y = x^4-8x^3-49x^2+260x+300. Setting y to zero and solving for x would give us all of the x-intercepts we see on the graph. However, this is a bit more algebra than we need for this article so we will save that equation for another day. Instead, if you need a few more examples, I have gone through some different types of functions in this video from the mathbootcamps youtube!



Finding x-intercepts on a Graphing Calculator

This is a bit more involved, but on this old youtube video I go through how to find both x and y intercepts using a TI 84.

What About y Intercepts?

Good question! I go over y-intercepts in this article.

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Matrix multiplication is just one of those things that is not intuitive – at least not at first. You have just had so many years of multiplication meaning one thing and then you have this entirely new definition to work with! It certainly takes some getting used to. (and if you continue to study advanced math – get used to that idea of “getting used to” things)

When I first studied matrix multiplication, I had an “aha!” moment that really helped. What was this?

Rows hit columns.

The first row of the first matrix hits every column and fill up rows in the new matrix. This process is repeated until you run out of rows in the first matrix. This is a bit difficult to imagine so I have created a small animation to help (there is no audio – just the animation):

This is really a way of visualizing the dot-product definition of matrix multiplication. For instance, when the first row of matrix 1 “hits” the first column of matrix 2, we see the sum 1(5)+2(7) results. This comes from the following definition:


\left[ \begin{array}{c}  a\\  b\\   \end{array} \right]  \left[ \begin{array}{cc}  c & d\\   \end{array} \right] = a(c)+b(d)

The next entry is therefore 1(6)+2(8). At this stage, you have run out of columns to “hit” and the next row in the first matrix is used.

As you continue to study matrices, you will likely find that this way of thinking about matrix multiplication works whether we are looking at multiplying two square matrices as above or more “odd” shaped matrices like those that can come up in linear algebra and similar courses.

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The phrase “two step equation” simply refers to linear equations that take two steps to solve. This comes from the idea of a one-step equation which, you guessed it – takes ONE step to solve! See, you got it already! Just to make it perfectly clear, let’s take a look at an example.

The equation 3x+5=14 is a two step equation. Why? In general, to solve any linear equation you want to isolate the variable. In other words, you want to get the variable by itself.

In order to do this, the first step you would take here is to subtract the 5 from both sides. Doing this will give you the equation 3x=9. Now, the only thing “happening” to the x is that it is being multiplied by 3. In order to undo this, you must divide by 3. Ahhh – the second step. Doing this gives you the wonderful equation x=3. Why is this wonderful? ‘cuz its the answer! Not only that, but you got it in TWO STEPS (did I drive that point home enough yet?).

Another example (without all the commentary – I promise) is 2x+6=24:

2x + 6 = 24
2x = 18 (subtract 6 from both sides)
x = 9 (divide by 2 on both sides).

In general, the steps to solve any “two-step” equation will be :

  1. Move the constant that is added or subtracted from the variable to the other side of the equation – If the number is added to the variable (like x), subtract it from both sides. If it is subtracted from the variable, add it to both sides.
  2. Divide both sides by the coefficient – The coefficient is the number that is multiplying the variable (often x). So, if you have a 5x, you will divide both sides by 5 at this step.

Once your equation is in the form “your variable = a constant (a number)”, you know you are done! Well, not completely… see, you aren’t really done until you have followed me on twitter. Doing that will make sure you don’t miss any new math tips that I post ;).

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While there are plenty of methods to get a good hand drawn graph of a linear equation, many plot more points than necessary and (if you aren’t careful) misrepresent the location of two important points: the x and y intercepts.

Using the intercepts to plot your line takes care of both of these problems and more importantly: its fast! In this video, I will use two examples to show you exactly how it’s done.

http://www.mathbootcamps.com/media/fastgraphsoflinearequations.mp4
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At this point, I haven’t even written out any equation or expression and you already know what I’m talking about. There are problems that come up in algebra where it seems like you should be able to cancel a variable out and move on. Yet, in these very situations your professor keeps telling you it can’t happen!

It may not always seem like it, but there really is a pattern and a simple rule of when you can and can’t cancel a term or a variable. Let’s focus on rational expressions specifically. Those are expressions (like the one below) where both the numerator and the denominator are polynomials.


\dfrac{x+5}{x^2+1}

When can you cancel terms in a rational expression?

You can cancel any matching factors that occur in both the numerator and the denominator. Let’s use numbers to understand this a little better:


\dfrac{4}{14}


In the fraction above, 2 is a factor of 4 since 2 x 2 is 4. Similarly, 2 is a factor of 14 since 2 x 7 = 14. Based on my rule, I should be able to cancel the 2’s and get a fraction that is still the same thing! Let’s see:



\dfrac{4}{14}=\dfrac{2(2)}{2(7)}=\dfrac{2}{7}

If you check on a calculator you will find that 4 divided by 14 is the same as 2 divided by 7 – so this rule seems to be working here. What is really going on?

Why does this work?

Every number other than zero, when divided by itself is 1. Since a fraction simply represents division (in one sense), \dfrac{2}{2} = 1 and we really just have 1\times\dfrac{2}{7}.

Trying it with Variables

Now that everything makes some sense with numbers, let’s try variables. For us, the variables represent numbers so they will behave in the same way. Take for instance the expression


  \dfrac{x^2-x}{x^3}, x\neq 0

The numerator has a couple of factors which we can find by factoring out an x (doesn’t that phrase make a lot more sense now?!): x^2-x = x(x-1). This shows us that x and x-1 are factors of the numerator. Looking at the denominator, we see x is also a factor since x^3=x(x^2). Remember, when both the numerator and the denominator share a factor, you can cancel that factor since it is really just 1.

  \dfrac{x^2-x}{x^3} = \dfrac{x(x^2-1)}{x(x^2)}= \dfrac{x^2-1}{x^2}

The only time you can’t cancel terms in the numerator and denominator is when they are both NOT factors.. That’s why you can’t cancel x in \dfrac{x-5}{x}. The x is not a factor of the numerator; its just a term being added. Cancelling the x here would be like cancelling the 5 in \dfrac{5+1}{5} and saying that is 1. If you check it in a calculator it certainly isn’t!

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There are many many ways to solve a quadratic equation (any equation that can be written as ax^2+bx+c=0), but today I’m going to focus on using the quadratic formula. Surely by now, you have seen a nice song to help you memorize it right? And you saw this nice tool to check your answer with right?

First the formula itself! When ax^2+bx+c=0 (quick notes: these are all real numbers and a is not zero) then

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
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Now, a, b, and c are all numbers from your equation. That means, as long as you can figure out what a,b,c are then you can just plug them into this equation and get the solution or solutions.

For example, take a look at this quadratic equation:

x^2-x-6=0.

In this example, a=1, b= -1, and c=-6. Why are b and c negative? The formula is based off the form ax^2+bx+c=0, so if you come across anything that is negative, keep the sign when you use the formula!

x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}

=\dfrac{1\pm\sqrt{1+24}}{2}

=\dfrac{1\pm\sqrt{25}}{2}.

At this stage, the plus or minus symbol (\pm) tells you that there are actually two different solutions:

\dfrac{1+\sqrt{25}}{2}=\dfrac{1+5}{2}=\dfrac{6}{2}=3 and

\dfrac{1- \sqrt{25}}{2}=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2.

Therefore the final answer would be x=3,-2.

That example worked out very nicely but sometimes there will be a bit more work to do. Take, for instance,

2x^2+2x-7=0

In this case, a=2,b=2, and c=-7. Therefore

x=\dfrac{-2\pm\sqrt{(-2)^2-4(2)(-7)}}{2(2)}

=\dfrac{-2\pm\sqrt{4+56}}{4}

=\dfrac{-2\pm\sqrt{60}}{4}

=\dfrac{-2\pm 2\sqrt{15}}{4}

Notice that 2 is a FACTOR of both the numerator and denominator, so it can be cancelled.

=\dfrac{-1\pm\sqrt{15}}{2}.

This answer can not be simplifed anymore without using decimals. Therefore the final answer is: x=\dfrac{-1+\sqrt{15}}{2},\dfrac{-1-\sqrt{15}}{2}.

It is also possible to discover that you have no solution or solutions that are no real numbers. We will take a look at those in a future post!

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While expressions with many exponents and variables can look complicated, simplifying them is really just about applying the laws of exponents one by one. When working on problems like these, make sure that you are never combining two terms with different bases and not leaving any negative exponents (this varies but it is most common to make sure all exponents are positive).

In this video I will go over four different types of examples and show you another way of thinking about these types of problems!

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To use this solver, your quadratic equation should be written in the form ax^2+bx+c=0. Simply input the coefficients (a,b,c) into the space below! While it will show you the steps, be aware that wolfram alpha (the program behind this widget) likes to complete the square for just about everything which may not be the most straightforward way to get your answer.

(I will probably add cool wolfram alpha widgets here over time but if you are looking for one for your class of website head on over this way: http://www.wolframalpha.com/widgets/

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