Find the Asymptotes f(x)=(x^2+2x-3)/(x^2+4x-5)

Problem

ƒ(x)=(x2+2*x−3)/(x2+4*x−5)

Solution

1. Factor the numerator and the denominator to identify potential discontinuities and simplify the function.

x2+2*x−3=(x+3)*(x−1)

x2+4*x−5=(x+5)*(x−1)

2. Simplify the expression by canceling common factors.

ƒ(x)=((x+3)*(x−1))/((x+5)*(x−1))

ƒ(x)=(x+3)/(x+5),x≠1

3. Identify the vertical asymptote by finding the values of x that make the denominator of the simplified function zero.

x+5=0

x=−5

4. Identify the hole (removable discontinuity) at x=1 because the factor (x−1) was canceled from both the numerator and denominator.

ƒ(1)=(1+3)/(1+5)=4/6=2/3

Hole at *(1,2/3)

5. Find the horizontal asymptote by comparing the degrees of the numerator and denominator or by taking the limit as x approaches infinity.

(lim_x→∞)((x2+2*x−3)/(x2+4*x−5))=1/1

y=1

Final Answer

Vertical Asymptote: *x=−5, Horizontal Asymptote: *y=1

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