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Graph a Rational FunctionIn this section, we will discuss graph a rational function.The rational function is of the form $f(x)=\frac{p(x)}{q(x)} where q(x)\neq 0$ The denominator of a rational function is never allowed to be zero; Since the division by zero is undefined. Steps to graph a rational function1) Reduce the rational function to lowest terms and check for any open holes in the graph. 2) Find x and y Intercepts of a rational function.3) Asymptotes. For vertical asymptote : Set the denominator equal to zero and solve. For horizontal asymptote : If the degree of numerator and denominator are same then there is horizontal asymptote. $y=\frac{(numerator′sleadingcoefficient)}{(denominator′sleadingcoefficient)}$ Slant or oblique asymptote : It can be found by long division. Remember the rational graph CANNOT cross the yaxis anywhere else and it CANNOT cross the xaxis at all! 4) Consider some values of x and according to that values find the values of y. Plot these points on the graph. Examples : For the following function, find a) xintercepts and yinercepts. b)vertical asymptotes. c) Horizontal/slant asymptotes. d)Holes e) Sketch the graph of the function. 1) $f(x)=\frac{2x}{x^21}$ Solution : $y =\frac{2x}{x^21}$ For xintercepts : Set y= 0 $0 = \frac{2x}{x^21}$ ∴ 2x = 0 (cross multiply) $\Rightarrow $ x = 0 So xintercept : (0,0) For yintercepts : Set x= 0 $y = \frac{2(0)}{0^21}$ $y = \frac{0}{0^21}$ $y = f(x)\frac{0}{1}$ ∴ y = 0 $\Rightarrow $ y = 0 So yintercept : (0,0) For Vertical asymptote : Set denominator equal to zero $x^{2}$  1 =0 $x^{2}$ = 1 ∴ x =$\pm$ 1 So vertical asymptote : x = 1 For horizontal asymptote : The degree of numerator is 1 and the degree of denominator is 2 so the margin is 1. ∴ Horizontal asymptote : y = 0 For Hole : Factor the denominator $x^{2}$  1 = (x + 1)(x  1) No common factor in the numerator. ∴ Hole : None 2) $f(x) =\frac{x^2+5x+6}{x^2+3x+2}$ Solution : $y =\frac{x^2+5x+6}{x^2+3x+2}$ For xintercepts : Set y= 0 $0 = \frac{x^2+5x+6}{x^2+3x+2}$ ∴ $x^{2}$+5x+6 = 0 (cross multiply) ∴ (x +3)(x+2) = 0 (factorize) So, x =  3, 2 So xintercept : (3,0) and (2,0) For yintercepts : Set x= 0 $y = \frac{0^2+5(0)+6}{0^2+3(0)+2}$ $y = \frac{6}{2}$ ∴ y = 3 So yintercept : (0,3) For Vertical asymptote : Set denominator equal to zero $x^{2}$+3x+2 = 0 (x+2)(x+1) = 0 (after factoring) x + 2 = 0 $\Rightarrow $ x = 2 x + 1 = 0 $\Rightarrow $ x = 1 So vertical asymptote : x = 2,1 For horizontal asymptote : The degree of numerator and the degree of denominator are same so the horizontal asymptote y = 1/1 ∴ Horizontal asymptote : y = 1 For Hole : Factor the denominator numerator and denominator $y =\frac{(x+3)(x+2)}{(x+2)(x+1)}$ common factor in the numerator and denominator is (x + 2) ∴ Hole : x = 2 precalculus Home Covid19 has led the world to go through a phenomenal transition . Elearning is the future today. Stay Home , Stay Safe and keep learning!!! Covid19 has affected physical interactions between people. Don't let it affect your learning.
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