Shifting graphs horizontally (also known as horizontal translation) is slightly different from vertical translation, but still pretty straight-forward. Perhaps it would be helpful to review my posting on vertical shifts of graphs. Recall from that section: “Picture all the complex stuff that is happening to x as being one chunk of the height component, and then when you add the + 5 to the equation, you are really just adding an additional height chunk to the total height for a given x.” I think this simplification condenses the rest of that post down quite nicely.

Shifting Graphs – Horizontal Translation

Now, to shift a graph horizontally, you include the shift amount with x. So, whatever action was being done just to x before, now you do that same thing to x plus the shift amount. Make sense? Probably not.

Check out the example below that hopefully demonstrates this better than I can explain with words.

If you want to shift the original function of f(x) = x² + 4 by 3 units, it becomes f(x) = (x-3)² + 4.

Can you see what I mean by including the shift amount WITH x. The ‘square’ function acts on the entire (x-3) term. This will cause the graph to shift 3 units to the RIGHT. This may seem somewhat counter-intuitive, but it is correct. Subtracting terms from x shift the graph to the right, whereas adding terms to x will translate them to the left.

In this example, x-3 causes a horizontal translation of the graph 3 units right… if it were x+3, it would translate the graph 3 units left. Here is a bit of a trick you can use to help you recall the direction of the shift caused by the signs. It may be easier to remember this by analyzing the “x and shift amount”, letting this small term equal to 0, and then solving for x. The result will show you how many units to move, and in what direction. Like this:

x – 3 = 0

x = 3 (shift 3 units right)

OR

x + 3 = 0

x = (-3) (shift 3 units left)

That shows you how far over, and in what direction, the new x values are! Technically, this is a way of finding a zero of the graph, but that is another post for another day. For now, I think it’s a helpful trick to apply at this stage!

I hope these postings on graph manipulations are helpful. Horizontal translation of functions and their graphs is still quite simple, albeit with the trick with the signs that you don’t have to worry about with vertical translation.

ALSO READ: Using the Quadratic Formula

In addition to the three basic trig functions we’ve already looked at (Sine, Cosine, Tangent), there are three other related functions. These are Secant, Cosecant, and Cotangent. These functions have similar meanings as the first three, in that they represent the ratios of various side lengths of a right angle triangle, and can be used to find angles or unknown side lengths. I will not go into extensive detail on these functions, as they are less commonly required, but I will show you what they mean.  Please remember to click on the Like button if this is helpful, and be sure to follow me with the buttons on the right side!

So far, with the help of SOHCAHTOA, we have seen that:

Sine = opposite / hypotenuse

Cosine = adjacent / hypotenuse

Tangent = opposite / adjacent

These new functions are related to the originals because they represent the inverse ratios. (Of course, inverse means you swap the top and the bottom…)

Cosecant = hypotenuse / opposite… (compare to Sine)

Secant = hypotenuse / adjacent…… (compare to Cosine)

Cotangent = adjacent / opposite…… (compare to Tangent)

Also, these functions can be abbreviated: Cosecant = Csc, Secant = Sec, Cotangent = Cot.

At the middle school and high school math level, you will rarely have a need to use these functions, but it is good for you to know what they are. However, most trig problems at this stage can easily be solved with the original three functions.  Just in case, though, it’s always good to know all the trig functions: sine, cosine, tangent, secant, cosecant, cotangent. If you’d be interested in additional material, a good place to start would be these links that explain secant, cosecant, and cotangent. You can also search that site for many other math concept explanations (just remember to come back here!)

Thanks for reading this quick “cheat sheet” version of these additional trig functions. Please remember to bookmark my site to come back again!

ALSO READ: Why Does m Represent Slope?

So, now that you know the answer to ‘What is the Quadratic Formula’ next I will show you examples of using it. Refer back to my last post to familiarize yourself with what the quadratic formula looks like. I’ve also explained there the nature of the roots of a quadratic equation. If you haven’t read it, I recommend taking a look as it might help you to visualize and to find the solution to a quadratic equation easier.

For my first example of using the quadratic formula to find the roots of a quadratic equation, let’s keep it simple.

x² – 2x –3 = 0

Comparing this to the standard form of a quadratic equation, ax² + bx + c = 0, we can equate the letter coefficients to the values provided. That is, we can say that a = 1, b = (–2), c = (–3). Now, we can simply substitute these values into the quadratic formula:

So, we have:

If you follow along with the arithmetic, you can see that we’ve solved the quadratic formula to show that the roots of the given equation are x = 3 and x = (–1).

Now, remember that I said in a previous lesson that you have to check your answers! Substitute these values back into the original equation, and you will find that they do indeed satisfy the equation. So, these are the correct roots!

Of course, you may have noticed that this question didn’t actually require the quadratic formula to solve for the roots. The quadratic formula worked well and got us the answer, but as you saw, it required a bit of work. And more work means more opportunity to make a mistake! You may have noticed that there was actually a faster way of solving the question. If you noticed that you could reduce the question down to (x – 3)(x + 1) = 0, you could simply let each set of brackets equal zero, and then find again that x = 3 and x = (–1) are the correct solutions.

Let’s try another one, adding some more of the previous math concepts I’ve gone over.

Using the quadratic formula, find the roots of:

2x³ + 3x² = 4x

It’s looks a little more complicated than the last one, huh? It has higher order exponents, and it doesn’t immediately look like a quadratic equation, as the first example did. However, with a little bit of arithmetic, and using your skills from the math concepts I explained in my post about factoring, it will begin to look a bit more familiar and workable.

So then, apply grouping techniques to our question. Let’s bring everything to one side first though. Recall that the standard form of a quadratic equation equals zero.

2x³ + 3x² = 4x
2x³ + 3x² – 4x = 0
x(2x² + 3x – 4) = 0

Looks a little better now, right? Maybe, something that might fit into the quadratic formula? Recall that the roots, or solutions, are any values of x that make the expression true. So, what we have derived up to this point is a product of two expressions that equals zero, and therefore the roots will be whatever values of x cause each part of the product to equal zero. The first (potential) root is obvious, from the first of the two expressions in the product: x = 0. (Substitute 0 back into the original equation to verify this is a correct root!) The second part, 2x² + 3x – 4, will require more work, and if we let it equal zero, you can see that it will fit into the quadratic formula perfectly.

To prepare for the quadratic formula, we need to identify our a, b, and c values. They are: a = 2, b = 3, and c = (–4). Now, we just substitute into the formula, do the math, and come up with our root(s) for this part of the question!

So, these are our answers for the two roots to the quadratic expression part of our original question. These are the radical forms of the solutions, so they look way more complicated. But, often the quadratic formula doesn’t reduce all the way down to a nice, round number and you will be left with something like this. The last thing you have to do is substitute them back into the original question to verify the roots are true, and that is it! Of course, when you write your answers down, make sure you remember to include the roots from the first part of the question, i.e. the part we created by grouping and solved for x = 0.

That last question goes over a lot of math concepts and is definitely comparable to some of the more complicated math questions you may find in your homework or on exams. Review and study it and make sure you understand it. I’ll post another example as well soon, if anyone needs some more examples of using the quadratic formula.