Monday, December 7, 2015
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Wednesday, December 2, 2015
Cameron Rational function graphing quiz
To begin with graphing, both equations should be factored in order to simplify them. Afterwards the vertical asymptotes should be identified. The vertical asymptotes require that,you determine when the denominator equals 0. So you find the number or numbers that when input for X equals zero . Then you can determine if the asymptotes are removable. In the case of the first one the asymptote is removable because x+1 over x+1 can be divided away. This changes the graph entirely. The equation will then become y=x-4 although there will be a hole in the equation and to find out the hole plug in the asymptote previously found for the equation before removing it. So you take the X and plug it into the equation and then you can find the y value and will have your hole . Then graph as you would a linear equation; just keep in mind that there will be a gap in the graph at the coordinates. The other equation does not have a removable asymptote so we'll take some different steps.Before anything the horizontal asymptote for this graph is 0 because the numerator has the same degree as the denominator. After the first thing that you should do when trying to graph a reciprocal function is find the intercepts. It makes plotting points on the graph so much easier. To do so you find when y is equal to 0 for the x-intercept and you find when x is equal to 0 for the y-intercept. Then you can begin by plugging on points to make the graph. Keep in mind that the graph will have two or three different sections due to the asymptotes so be sure to account for all spaces on the graph. This completes your graphing experience if you have any need of reference refer to the guide attached with numbered steps to each equation.
Ben Emerson rational graphing quiz
If you were to attempt graphing the 2 rational functions in the picture shown below, you must carefully go through the steps of the graph before you can actually graph your equation. The first equation is called a removable discontinuity meaning you can factor, and cancel 2 of the same factors on the numerator, and denominator. This makes a hole in the graph at that point. The vertical asymptote is -1 because if you plug it in for X the numerator is 0. The horizontal asymptote is non existent because the numerator has a higher degree than the denominator. The y intercept can be found by plugging in a 0 for x and then simply solving. The x intercept is what ever makes the numerator 0, in this case it be 4. Lastly a table of values can really help plot the points on the graph, but don't forget your vertical asymptote can't be an X value!
The second function has different steps to solving. It does not have any holes because nothing is factored out, use all the same steps to find your vertical, horizontal asymptotes, as well as your x, and y intercepts. The graph looks a little funky also, see graph below for questions.
Mike C Rational function blogging quiz
The difference between graphing the two equations is the first equation is a removable discontinuity and the second graph is a non removable discontinuity. To graph the first equation I first factored the top of the equation. After that I realized I was able to simplify the equation by canceling out the two x+1s. This creates a hole at -1 because -1+1 would be zero. After that I found the vertical asymptote a which would be -1 and 4, because 4-4=0. Since the degree of the numerator is greater than that of the denominator, there would be no horizontal asymptote. To make the numerator 0, you would have to plug in a 4 for x, that is where I got my x intercept. When you plug in 0 for x you get -4, that is where I got my y intercept. I then used all of these features to grt my graph laid out. I then created a table of x values within the asymptotes, and solved them to get the y values. Then I graphed the points.
For the second graph I started out the same way, except this time it was the denominator that needed to be factored. This time nothing needed to cancel out after I factored. The vertical asymptotes would be 2 and -2 since 2-2=0 and -2+2=0. The degrees are the same in this equation and 1/1 is 1, that's where the horizontal asymptote comes from. The intercept is 4 because 4-4=0 in the numerator. The y intercept ends up being 0 when you plug in 0 for x. Then I laid out the graph just as I did with the last equation and plotted the points the same way.
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