Question

Sketch of y = (x^2 - x + 1)/(x^2 + x + 1) with key features

Original question: (d) Sketch the curve with equation $y = \frac{x^2-x+1}{x^2+x+1}$.

Expert Verified Solution

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Key takeaway: This rational function has a nice structure, and it simplifies the sketch a lot once you compare numerator and denominator. The key is to look at end behavior first, then test for intercepts and any symmetry.

Consider

y=x2βˆ’x+1x2+x+1.y=\frac{x^2-x+1}{x^2+x+1}.

1) Horizontal asymptote

The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients:

y=1.y=1.

2) Intercepts

  • yy-intercept: set x=0x=0
y=11=1,y=\frac{1}{1}=1,

so the graph passes through (0,1)(0,1).

  • xx-intercepts: solve
x2βˆ’x+1=0.x^2-x+1=0.

Its discriminant is

(βˆ’1)2βˆ’4(1)(1)=1βˆ’4=βˆ’3<0,(-1)^2-4(1)(1)=1-4=-3<0,

so there are no real xx-intercepts.

3) Compare with the asymptote

Subtract 1:

yβˆ’1=x2βˆ’x+1βˆ’(x2+x+1)x2+x+1=βˆ’2xx2+x+1.y-1=\frac{x^2-x+1-(x^2+x+1)}{x^2+x+1} =\frac{-2x}{x^2+x+1}.

Since

x2+x+1>0forΒ allΒ realΒ xx^2+x+1>0 \quad \text{for all real }x

because its discriminant is also negative,

the sign of yβˆ’1y-1 is the sign of βˆ’x-x.

So:

  • if x<0x<0, then y>1y>1
  • if x>0x>0, then y<1y<1
  • if x=0x=0, then y=1y=1

That tells you the curve crosses the horizontal asymptote at (0,1)(0,1).

4) Shape

The curve stays defined for all real xx and never hits the xx-axis. It lies above y=1y=1 for negative xx and below y=1y=1 for positive xx, with a crossing at (0,1)(0,1).

If you want a quick extra check, evaluate a couple of values:

y(1)=1βˆ’1+11+1+1=13,y(1)=\frac{1-1+1}{1+1+1}=\frac13, y(βˆ’1)=1+1+11βˆ’1+1=3.y(-1)=\frac{1+1+1}{1-1+1}=3.

So the sketch passes through (βˆ’1,3)(-1,3) and (1,13)(1,\tfrac13), approaching y=1y=1 on both ends.


Pitfalls the pros know πŸ‘‡ A frequent mistake is to assume the horizontal asymptote cannot be crossed. That is false here: the graph does cross y=1y=1 at x=0x=0. Another slip is forgetting to check whether the denominator can be zero; here it cannot, so there are no vertical asymptotes.

What if the problem changes? If the signs in the denominator changed to x2βˆ’x+1x^2-x+1, then the curve would still have horizontal asymptote y=1y=1, but the comparison with 1 would change. Tiny sign changes in rational functions often alter where the graph lies above or below the asymptote, even when the asymptote itself stays the same.

Tags: horizontal asymptote, rational function, curve sketching

FAQ

What is the horizontal asymptote of (x^2-x+1)/(x^2+x+1)?

Since the numerator and denominator have the same degree, the horizontal asymptote is y=1.

Does the graph cross its horizontal asymptote?

Yes. Subtracting 1 gives (y-1)= -2x/(x^2+x+1), so the graph crosses y=1 at x=0.

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