# Graph a Rational Function

In 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 function**

1) 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 y-axis anywhere else and it CANNOT cross the x-axis 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) x-intercepts and y-inercepts.

b)vertical asymptotes.

c) Horizontal/slant asymptotes.

d)Holes

e) Sketch the graph of the function.

1) $f(x)=\frac{2x}{x^2-1}$

**Solution :**$y =\frac{2x}{x^2-1}$

**For x-intercepts :**Set y= 0

$0 = \frac{2x}{x^2-1}$

∴ 2x = 0 (cross multiply)

$\Rightarrow $ x = 0

So x-intercept : (0,0)

**For y-intercepts :**Set x= 0

$y = \frac{2(0)}{0^2-1}$

$y = \frac{0}{0^2-1}$

$y = f(x)\frac{0}{-1}$

∴ y = 0

$\Rightarrow $ y = 0

So y-intercept : (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 x-intercepts :**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 x-intercept : (-3,0) and (-2,0)

**For y-intercepts :**Set x= 0

$y = \frac{0^2+5(0)+6}{0^2+3(0)+2}$

$y = \frac{6}{2}$

∴ y = 3

So y-intercept : (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

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