Wednesday, May 24

In class we worked on converting functions into rational form.

Assignment: 9.8 on p. 555: 1-8, 11-12

Here is an example problem, similar to the one worked in class.

Let's analyze the function given by the equation below. We will try to work out its various characteristics and sketch a graph before resorting to computer graphing.

y = dfrac{x-1}{(x+1)(x - 2)} + dfrac{2x - 1}{x^2 - 4}

To obtain a common denominator we multiply the first term by (x+2)/(x+2) and the second term by (x + 1)/(x+1)

y = dfrac{(x - 1)(x + 2)}{(x + 1)(x - 2)(x + 2)} + dfrac{(2x - 1)(x + 1)}{(x + 1)(x - 2)(x + 2)}

Now we can simplify the expressions in the numerators and re-write as a single fraction.

y = dfrac{x^2 + x - 2 + 2x^2 + x - 1}{(x +1)(x - 2)(x + 2)}

Which simplifies to

y = dfrac{3x^2 + 2x - 3}{(x + 1)(x - 2)(x + 2)} or y = dfrac{3x^2 + 2x - 3}{x^3 + x^2 - 4x - 4}

We note that 3x^2 + 2x - 3, the numerator, cannot be factored. So, later on as we analyze the properties of this function we will need to solve 3x^2 + 2x - 3 = 0 to determine the value of any x-intercepts.

Now that the function is in rational form, we can analyze its properties.

We note that there are no holes in the function. Because there are no factors common to both the numerator and denominator. i.e. (x + 2)/(x + 2). These common factors would work out to essentially multiplying the function by 1 for all values of $x$ except for when $x + 2 = 0$ at $x = -2$. For that single value $(x + 2)/(x + 2)$ becomes undefined and we get a hole in the graph. Anyway, since this is not the case, our function has no holes.

The function has vertical asymptotes at

x = -1, x = 2 and x = -2

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We can tell from the numerator that the function will have x-intercepts where the numerator evaluates to zero, so we solve 3x^2 + 2x - 3 = 0 to get

x = dfrac{-2 pm sqrt{2^2 - 4(3)(-3)}}{2(3)}

x = dfrac{-2 pm sqrt{40}}{6} = dfrac{-2 pm 2sqrt{10}}{6} = dfrac{-1 pm sqrt{10}}{3}  approx -1.387,0.72

So, we note the two x-intercepts.

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In addition, we note that this function has denominator of greater degree than the numerator, so as x approaches pm infty the denominator will overpower the numerator and the value of the function will approach 0, creating horizontal asymptotes at y = 0

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Through point testing and some more intuition we can identify the location of the 4 sections of this function and sketch a graph

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