Sunday, January 27, 2013

On Thursday in class we started to learn about y intercepts on graphs of functions. Here is a graph we looked at:
Equation: y = (x+7) (x+1) (x-6) (x-10)^2
y intercept : -4200
Though we can't see it this picture of the graphed function (the window would have to be huge!!) the y - intercept of this function is -4200.
After graphing this with our calculators in class and playing around with our windows for a while, Lisa asked us to now change the y - intercept to -42. At first I was like, "what?!" but then I realized that a few weeks ago we learned about reflection over the axes, vertical shift, vertical stretch, vertical shrink, horizontal shift, horizontal stretch, and horizontal shrink. Changing the y-intercept of this function from -4200 to -42 is a vertical shrink. 
But how do you change the y-intercept?
Like this: 
y = (x+7) (x+1) (x-6) (x-10)^2 (.01)
By multiplying our old equation by (.01), the function will undergo a vertical shrink.
Now the function looks like this:



Equation: y = (x+7) (x+1) (x-6) (x-10)^2 (.01)
y- intercept: -42
Let's try something else, like a reflection over the x axis keeping our vertical shrink. 
Take the equation of the function:
y = (x+7) (x+1) (x-6) (x-10)^2 (-.01)
By making the (.01) negative, the function is now the reflection of what it used to be. 
Here's what it looks like now:


Equation: y = (x+7) (x+1) (x-6) (x-10)^2 (-.01)
y- intercept: 42

See the difference?
On Friday, we talked about x intercepts, and how do find their exact location on a graph (without even really needing a calculator!) Sometimes x intercepts are easy to see with our eyes:


The function is clearly going through the x axis at (-2,0) (-1,0) and (2,0). Therefore, it is easy to find an equation to this function:
y = (x+2) (x+1) (x-2)
The +2, +1, and -2 are the negative version of the x - intercepts on the graph - 2, - 1, and 2.
However, sometimes it is hard to tell where exactly the function is hitting and/or going through the x axis (I say and/or because remember - if the x is squared then it will be a parabola on the axis instead of going through.)
Let's take this confusing function for example:


Again, we can clearly see that the function goes through the x axis as -3, but WHERE does it go through in the in the other two places??? It goes through somewhere between -1 and -2, and then again somewhere between 1 and 2. But where exactly?
Here's how to find out:
This is the equation of the function: f(x) = 
1. Factor the equation 
2. Factor it completely  y = (x+3) (x + route2) (x - route2)
3. This means that - 3, - route2 and positive route 2 are our exact x intercepts.

Then we learned how to graph functions on our own by picking apart an equation with what we know and how we can use synthetic division to locate points of the function on a graph.
Let's take this equation: (x + 1)^3 (x) (x-1)
1. By looking at this equation, we can see that this is a 5th degree function because there are five xs total. This means that the function will start low, because it is odd.
SIDE NOTE: a function with an odd number of xs starts low, and a function with an even number of xs starts high.
2. Now we need to find the x intercepts in order to graph it without a calculator. x-intercepts: -1, 0, and 1.
3. Always pay attention to x^2, x^3, x^4s and so on when graphing. In this function in particular, we have an x^3, meaning that the function WILL NOT look like this:


The function is not this simple on the graph. The function pictured here is (x+1) (x) (x-1) not 
(x+1)^3 (x) (x-1)
SIDE NOTE: When we have an x^2: 

When we have an x^3:


When we have an x^4:
When we have an x^5:

etc.

So, with this in mind, we would graph our equation (x + 1)^3 (x) (x-1) like this:



Using Synthetic Division to Locate Exact x-intercepts!
Lastly, I would like to go over synthetic division using the homework problem we had for Monday as an example.
Function Equation:
f(x) = x^4 - 6x^3 + 4x^2 + 15x + 4
Step 1. Plug this equation into your calculator. You will get a graph that looks like this:


See how the function goes through the axis 4 times? That is because it is a fourth degree polynomial, its leading exponent is 4. In 2/4 places we can see exactly where these x - intercepts are - one is at (-1,0) and the other one is at (4,0). 
Chose either one of these x-intercepts for your synthetic division.
Step 2. I'm going to use the -1 x-intercept for my factoring. Then, I take all of the coefficients of our equation f(x) = x^4 - 6x^3 + 4x^2 + 15x + 4 and divide them by negative one using synthetic division


Step 3. now I'm going to run these coefficients through synthetic division once again until it is all factored, still dividing by negative one: 













Now, our equation looks like this: y = (x+1)(x-4) (1x^2 -3x -1)

Step 4: To find those two other x-intercepts, we are going to factor the remaining un-factored part of our equation using the quadratic formula to find x. 



Those are your two other x-intercepts!











6 comments:

  1. This post was very helpful to me because in class when we were doing this I was somewhat confused but this really helped me to understand it better. I like how you used a lot of pictures to explain the material, and the use of different colors made it more fun! Great job!!

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  2. WOAH this is quite impressive. I like the pictures and the descriptions! very informative!

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  3. Identifying the zeroes of an equation can be somewhat confusing, but having to do transformations on top of that makes for some challenging problems. You did a great job in this post of explaining and providing examples of how transformations and synthetic division will help you find where the function crosses the x axis.

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  4. Wow, Talia, this post is impressive! You've thoroughly covered the many connections between equations of polynomials and their graphical characteristics. The post reads easily and includes some nice descriptions and extensions of the concepts. Also, your graphs and other images look great and are essential for "seeing" the connections you discuss. (Just so you know, two of the equations did not appear, so you may want to go back and see if there is a way to edit this.)

    On a separate note, I must have never actually written the word "root" on the board. You have it written as "route" in your post. This is an oversight on my part, not yours.... I just wanted to make sure you knew.

    Nicely done, Talia! This would be a great post for others to read as they prepare for the test.

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  5. This was a very helpful post. I used it to review for the test and it really helped me understand some topics I wasn't confident with.

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  6. This was a very helpful post. I used it to review for the test and it really helped me understand some topics I wasn't confident with.

    ReplyDelete