Graph f(x)=(x^2-8)/(x+3)

Problem

ƒ(x)=(x2−8)/(x+3)

Solution

1. Identify the domain by finding where the denominator is zero.

x+3=0⇒x=−3

The domain is all real numbers except x=−3

2. Find the vertical asymptote by checking the behavior near the excluded value.
   Since the numerator x2−8 is not zero at x=−3 there is a vertical asymptote at x=−3

3. Find the slant asymptote using polynomial long division because the degree of the numerator is exactly one higher than the degree of the denominator.

(x2−8)/(x+3)=x−3+1/(x+3)

The slant asymptote is y=x−3

4. Determine the intercepts by setting x=0 and ƒ(x)=0
   For the yintercept:

ƒ(0)=(0−8)/(0+3)=−8/3

For the xintercepts:

x2−8=0⇒x=±√(,8)≈±2.83

5. Find critical points by calculating the derivative d(ƒ(x))/d(x) using the quotient rule.

d(ƒ(x))/d(x)=((x+3)*(2*x)−(x2−8)*(1))/((x+3)2)

d(ƒ(x))/d(x)=(x2+6*x+8)/((x+3)2)

Setting the numerator to zero:

x2+6*x+8=0⇒(x+4)*(x+2)=0

Critical points are at x=−4 and x=−2

6. Evaluate the function at critical points to find local extrema.

ƒ*(−4)=((−4)2−8)/(−4+3)=8/(−1)=−8

ƒ*(−2)=((−2)2−8)/(−2+3)=(−4)/1=−4

There is a local maximum at (−4,−8) and a local minimum at (−2,−4)

Final Answer

To graph ƒ(x)=(x2−8)/(x+3) plot the vertical asymptote x=−3 the slant asymptote y=x−3 the intercepts (0,−2.67) (2.83,0) (−2.83,0) and the relative extrema (−4,−8) and (−2,−4)

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