1FGraphingPartFracs


 * Graphs of Rational Functions Using Partial Fractions**

Use long division if the order of the numerator is greater than or equal to the order of the denominator.

Then use partial fractions if the denominator is a quadratic (or greater) that can be factorised.

Example math . \qquad y = \dfrac {-x^2 - 3}{(x+1)(x-3)} math


 * The order of the numerator = order of denominator so do long division. **
 * The denominator is: ** **//x 2 – 2x – 3//**

math . \qquad x^2-2x-3 \; \overset{\displaystyle {-1} \qquad \qquad .} {\overline { \left) -x^2+0x-3 \right. }} math math . \qquad \qquad \qquad \quad \; \underline {-x^2+2x+3} math math . \qquad \qquad \qquad \quad \qquad -2x-6 math

math \text {Thus } \; y = \dfrac{-2x-6}{(x+1)(x-3)}-1 math


 * ** Now use partial fractions on the fraction term  **

math \\ . \qquad \dfrac {-2x - 6}{(x+1)(x-3)} = \dfrac {A}{x+1} + \dfrac {B}{x-3} \\. \\ . \qquad \dfrac {-2x - 6}{(x+1)(x-3)} = \dfrac {A(x-3) + B(x+1)}{(x+1)(x-3)} math


 * ** Equate numerators  **

math . \qquad -2x - 6 = A(x-3) + B(x+1) math

math \\ . \qquad \text {Let } x = 3 \Longrightarrow -12 = A(0) + B(4) \Longrightarrow B = -3 \\. \\ . \qquad \text {Let } x = -1 \Longrightarrow -4 = A(-4) + B(0) \Longrightarrow A = 1 math

math \\ . \qquad \text {Thus } \dfrac {-2x - 6}{(x+1)(x-3)} = \dfrac {1}{x+1} + \dfrac {-3}{x-3} \\. \\ \\ . \\ . \qquad \text {So } \: y = \dfrac {1}{x+1} + \dfrac {-3}{x-3} - 1 math


 * The graph will have asymptotes at x = –1, x = 3, y = –1


 * Because we are adding 2 hyperbolas ,
 * the graph will cross the y = –1 horizontal asymptote when:

math \\ . \qquad \dfrac {1}{x+1} + \dfrac {-3}{x-3} = 0 \\. \\ . \qquad \dfrac {1}{x+1} = \dfrac {3}{x-3} \\. \\ . \qquad x - 3 = 3(x + 1) \\. \\ . \qquad x - 3 = 3x + 3 \\. \\ . \qquad 2x = -6 \\ .\\ . \qquad x = -3 math

To sketch this, first sketch the individual hyperbolas, Then sketch the final graph using addition of ordinates

Asymptotes x = –1, x = 3, y = –1

math \text {Domain: } x \in R \backslash \lbrace -1, 3 \rbrace math

Range is ?? {see below}

x-intercept none y-intercept y = 1

Cuts the horizontal asymptote at (–3, –1)

Notice that as x approaches negative infinity, y approaches the horizontal asymptote at y = –1 from __above__.

By observation of the graph, there is a local minimum between x = 0 and x = 1 and a local maximum between x = –10 and x = –3

Using the calculator, we get turning points at (0.464, 0.866) and (-6.464, -0.866)

math \text {So Range is } \lbrace y \leqslant -0.866 \rbrace \cup \lbrace y \geqslant 0.866 \rbrace math

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