Rational functions and graphs
Introduced| English | Meaning |
|---|---|
| polynomial/ˌpɒlɪˈnəʊmɪəl/ | An expression formed from a finite sum of non-negative integer powers of a variable, each multiplied by a coefficient. |
| rational function/ˈræʃənl ˈfʌŋkʃn/ | A function expressible as a ratio of two polynomials, with a non-zero denominator. |
| denominator/dɪˈnɒmɪneɪtə/ | The lower part of a fraction, indicating the divisor or the number of equal parts in a whole. |
| vertical asymptote/ˈvɜːtɪkl ˈæsɪmptəʊt/ | A vertical line that a graph approaches as its function values grow without bound near a particular input. |
| oblique asymptote/əˈbliːk ˈæsɪmptəʊt/ | A non-horizontal straight line that a graph approaches, with its vertical difference from the line tending to zero as x tends to an appropriate infinity. |
| quotient/ˈkwəʊʃənt/ | The result of dividing one quantity or expression by another. |
| remainder/rɪˈmeɪndə/ | The amount left after division into equal whole groups, or the residual expression after algebraic division. |
| turning points/ˈtɜːnɪŋ pɔɪnts/ | Points where a graph changes from rising to falling, or from falling to rising. |
| discriminant/dɪˈskrɪmɪnənt/ | The value b squared minus 4ac, which determines the nature of a quadratic equation's roots. |
Check a fitted curve before trusting it
- A program may fit a ratio of two polynomials 多项式 to data. Its graph can look smooth on a screen while hiding an excluded input or a separate branch.
- For a rational function 有理函数 $f(x)=P(x)/Q(x)$, begin with the original denominator 分母: require $Q(x)\ne0$. Then find the features that a sketch must show.
An excluded input can be a hole
- A denominator zero gives a vertical asymptote 垂直渐近线 only when the remaining expression becomes unbounded there. Factor first; cancellation may instead leave a missing point.
- For $f(x)=(x^2-1)/(x-1)=x+1$, the original restriction $x\ne1$ remains. Draw the line with an open point at $(1,2)$, not a vertical asymptote. A zero of the numerator is an intercept only if the original denominator is non-zero there.
A rational function is a fraction of two polynomials.
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That is the definition — one polynomial over another.
Sketch $(x^2-4)/(x-2)$ with its original domain. State whether x=2 is a vertical asymptote, and list its actual axis intersections.
The expression simplifies to x+2 only for x≠2. Draw that line with a hole at (2,4). The function is bounded near 2, so there is no vertical asymptote there. The actual intersections are (−2,0) and (0,2).
Asymptotes from degree and division
- After cancellation, compare degrees: a lower-degree numerator gives horizontal limit 0; equal degrees give the ratio of leading coefficients. An oblique asymptote 斜渐近线 is a sloping line, so the numerator degree must be exactly one higher.
- Divide to obtain the quotient 商 plus the remainder 余数 over the denominator. For $(x^2-1)/(x-2)$, this is $x+2+3/(x-2)$. Thus $y=x+2$ is the slant limit as $x\to\pm\infty$; $x=2$ is vertical.
A horizontal or slant asymptote describes distant behaviour; it may be crossed at a finite input. For $f(x)=(x^2+x)/(x^2+1)$, the horizontal asymptote is $y=1$, and $f(1)=1$.
For a reduced rational function with a non-zero remainder, a sloping straight-line asymptote occurs when the numerator degree is:
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When the numerator's degree exceeds the denominator's, dividing out leaves a slant line the curve approaches.
You find an oblique asymptote by:
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Polynomial division gives the straight-line part the curve approaches far from the origin.
For f(x) = (x²−1)/(x−2), divide to get f(x) = x + 2 + 3/(x−2). The oblique asymptote is y = x + ?
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The quotient is x + 2, so the oblique asymptote is y = x + 2.
Turning points 极值点 and a complete sketch
- For $f(x)=x+1/x$, $x\ne0$, $f'(x)=1-1/x^2$. The turning points are $(-1,-2)$, a local maximum, and $(1,2)$, a local minimum, as the derivative changes sign.
- There are no axis intersections. Draw both branches with vertical asymptote $x=0$ and slant asymptote $y=x$. The left branch lies below the slant line and the right above it. Near zero, the left branch falls to $-\infty$ and the right rises to $+\infty$.
Not every rational curve has stationary points. For $h(x)=(x+1)/(x^2+2x-2)$,
at every allowed input. Its numerator cannot be zero, so it has no stationary points and decreases on each interval of its domain. Do not join decreasing branches across a pole.
For $f(x)=(x^2+4)/(x-2)$, find the asymptotes, stationary coordinates, axis intersections and range, and draw a labelled sketch.
$f=x+2+8/(x-2)$, x≠2. Asymptotes x=2 and y=x+2. $f\prime=1-8/(x-2)^2$ gives x=2±2√2 and y=4±4√2; the left point is a maximum, the right a minimum. There are no real x-intercepts; the y-intercept is (0,−2). Range $(-\infty,4-4\sqrt2]\cup[4+4\sqrt2,\infty)$. Left branch lies below the slant line, right above.
For $h(x)=(2x+1)/(x^2+x-3)$, prove that there are no stationary points. State the separate intervals on which it decreases.
The original exclusions are $l=(-1-\sqrt{13})/2$ and $r=(-1+\sqrt{13})/2$. The derivative is $-(2x^2+2x+7)/(x^2+x-3)^2$. Its numerator before the minus sign is $2(x+1/2)^2+13/2>0$, while the denominator is positive on the domain. Thus the derivative is always negative and never zero. The function decreases separately on $(-\infty,l)$, $(l,r)$ and $(r,\infty)$.
Prove the range with a discriminant 判别式
- Rearrange $y=x+1/x$ as $x^2-yx+1=0$. A real input exists exactly when its discriminant satisfies $y^2-4\ge0$. The range is $y\le-2$ or $y\ge2$, including both turning values.
- Check exclusions and any case where the equation stops being quadratic. For $y=x/(x^2+1)$, rearrangement gives $yx^2-x+y=0$. At $y=0$ this is the valid linear equation $x=0$; do not discard that range value.
For the second function, $1-4y^2\ge0$ gives $-1/2\le y\le1/2$. Its denominator is never zero for real $x$, so all these values occur, including $0$ and the two endpoints.
Sketch $f(x)=(x^2-4)/(x^2-1)$ and find its range. Label poles, intercepts and its stationary point.
Domain x≠±1, vertical asymptotes x=±1, horizontal asymptote y=1. Intercepts (±2,0),(0,4). $f\prime=6x/(x^2-1)^2$, so (0,4) is a minimum on the middle branch. Put z=x²≥0, z≠1: $y=1-3/(z-1)$. For 0≤z<1, y≥4; for z>1, y<1. Range $(-\infty,1)\cup[4,\infty)$. All three branches must be shown.
Use a discriminant to find the range of $f(x)=(2x+1)/(x^2+1)$. Handle y=0 separately and identify the x-values where the range endpoints occur.
Rearrange to $yx^2-2x+(y-1)=0$. For y≠0, discriminant $4-4y(y-1)\ge0$ gives $(1-\sqrt5)/2\le y\le(1+\sqrt5)/2$. At y=0, x=−1/2 is valid. The denominator never vanishes. Equality gives a double input root x=1/y, so endpoint inputs are (−1−√5)/2 and (−1+√5)/2 respectively.
For a>0, sketch the features of $f_a(x)=(x^2+a^2)/(x-a)$: asymptotes, stationary coordinates and range. Explain what changes when a=0 in the original expression.
$f_a=x+a+2a^2/(x-a)$. Asymptotes x=a,y=x+a. Stationary points $(a\pm\sqrt2a,2a\pm2\sqrt2a)$; the minus point is a maximum and plus a minimum. Range $y\le2a-2\sqrt2a$ or $y\ge2a+2\sqrt2a$. At a=0 the expression is x²/x=x for x≠0: a line with a hole at (0,0), no vertical asymptote and no stationary points. Its range is R without {0}.
Change the output: modulus and reciprocal
- Take $f(x)=(x-1)/(x+1)$. For $y=|f(x)|$, keep nonnegative pieces and reflect negative pieces above the $x$-axis. The exclusion $x=-1$ remains. Its only zero is $x=1$.
- For $y=1/f(x)=(x+1)/(x-1)$, exclude both $x=-1$ and $x=1$: the first was never in the original domain and the second makes the reciprocal undefined. There is a hole at $(-1,0)$ and a vertical asymptote at $x=1$; the hole is not a zero.
For $f(x)=(x-2)/(x+3)$, sketch y=1/f(x) with the domain inherited from f. Give its poles, holes, zeros and range.
$1/f=(x+3)/(x-2)$ for x≠−3,2. There is a vertical asymptote at x=2, a hole at (−3,0), and horizontal asymptote y=1. There are no actual zeros. Since $y=1+5/(x-2)$ omits 1, and the only input giving y=0 is excluded, its range is R without {0,1}.
Change the input, or square the output
- For $f(|x|)$, keep the original right half and reflect it across the $y$-axis. Here the denominator $|x|+1$ never vanishes; the range is $[-1,1)$, with zeros at $x=\pm1$ and a corner at $(0,-1)$. Such a corner is not automatic: if the original right-hand slope at zero is zero, the join can be smooth.
- For $y^2=f(x)$, keep only original inputs where $f(x)\ge0$, then draw both $y=\sqrt{f(x)}$ and $y=-\sqrt{f(x)}$. Here $x<-1$ or $x\ge1$; the two branches meet at $(1,0)$, and no graph exists for $-1<x<1$.
A smooth join is possible: for $h(x)=(x^2+4x+8)/(x+2)$, $h(0)=4$ and $h'(0)=0$. Reflecting its right half gives matching slopes 0 on both sides, so $h(|x|)$ is smooth at $(0,4)$.
Explore a reciprocal curve
y = a/(x − b) + c
For a non-zero a, x=b is excluded and the curve approaches y=c at large |x|. Track the domain as well as the visible branches.
For y = 1/f(x), the zeros of f(x) become the zeros of y.
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A zero of f becomes an excluded input and usually a vertical asymptote of 1/f. Original domain exclusions remain; do not turn an original pole into a filled zero.
The graph of y = |f(x)| is obtained from y = f(x) by:
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The modulus reflects any part of the graph below the x-axis to above it.
For $f(x)=(x^2-4)/(x^2-1)$, sketch $y^2=f(x)$. State its real x-domain, axis intersections, vertical and horizontal limiting lines.
Require f≥0 in the original domain. This gives $x\le-2$ or $-1<x<1$ or $x\ge2$. Draw both signs of √f, symmetric about the x-axis (also even in x). Intersections (±2,0),(0,±2). Vertical asymptotes x=±1 approached from the central interval. As |x|→∞, the outer branches tend to y=±1. No points exist on (−2,−1) or (1,2).
For $g(x)=(x-2)/(x+1)$, sketch y=|g(x)| and y=g(|x|) separately. Compare domains, zeros, y-intercepts and the join at x=0. Would $h(|x|)$, for $h(x)=(x^2+4x+8)/(x+2)$, have the same type of join at zero?
For |g|, x≠−1, pole x=−1, sole zero x=2, y-intercept2; reflect original negative outputs upward. For g(|x|), the denominator |x|+1 is positive for every real x, zeros are ±2 and y-intercept−2. Copy only the right half of g and reflect left; the slopes at zero are −3 and +3, so there is a corner minimum there. Its range is [−2,1). For h, division gives $h=x+2+4/(x+2)$, so $h(0)=4$ and $h\prime(0)=1-4/4=0$. The reflected slopes both equal zero; its join at (0,4) is smooth, unlike the corner for g.
Solve inequalities from boundaries and branches
- To solve $|(x-1)/(x+1)|<2$, find where $f=2$ and $f=-2$: the boundary inputs are $-3$ and $-1/3$. Include the pole $-1$ in the sign or branch diagram.
- Test the intervals separated by these values. The answer is $x<-3$ or $x>-1/3$, with strict endpoints excluded. Do not multiply an inequality by $x+1$ without first deciding its sign; multiplying by $(x+1)^2$ is safe only after excluding $x=-1$.
Solve $|(x-2)/(x+1)|<1/2$ exactly, showing boundaries and the excluded input.
Exclude x=−1. The boundaries f=−1/2 and f=1/2 occur at x=1 and x=5. On x<−1, f>1, so there is no solution there. On x>−1, f increases from −∞ to1, giving 1<x<5. An equivalent safe algebra route squares and multiplies by the positive (x+1)² after excluding −1.
Use symmetry to solve an input-modulus inequality
- For $f(|x|)<1/2$, put $t=|x|\ge0$. Then $(t-1)/(t+1)<1/2$ has a positive denominator and gives $t<3$.
- Translate back to $-3<x<3$. The value zero is included. If a quadratic calculation produces a negative boundary for $t$, it cannot be a value of $|x|$; inspect the allowed half-line before reflecting the result.
Solve $((|x|)^2+4)/(|x|+2)<3$ exactly. Explain how the restriction on |x| affects the two quadratic boundary roots.
Put t=|x|≥0. Since t+2>0, the inequality becomes t²−3t−2<0. Its roots are (3±√17)/2, but the lower one is negative. Thus 0≤t<(3+√17)/2 and $-(3+\sqrt{17})/2<x<(3+\sqrt{17})/2$. Zero is included.
Build the sketch from evidence
- Record the domain, intercepts, asymptotes, stationary coordinates and one-sided behaviour. Use a range calculation to confirm which vertical gaps are real; a horizontal asymptote need not be an excluded output value.
- For a related graph, state the new domain before drawing. For an inequality, mark all boundaries and poles, test branches, and preserve strict or inclusive endpoints.
For (x squared plus 1) divided by (x minus 1), what is the oblique asymptote, and why is the remainder not that line?
Polynomial division writes the function as quotient plus remainder over divisor. Only the quotient survives as the oblique asymptote.
For (x²+1)/(x−1), what constant is added to x in its slant asymptote?
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The quotient is x plus 1, so the constant is 1. The remainder is 2.
Solve $|(x^2-4)/(x^2-1)|<1/2$ exactly. Give the intervals and show why multiplying by the denominator itself would require care.
Exclude x=±1. Squaring and multiplying by (x²−1)² gives $4(x^2-4)^2<(x^2-1)^2$, hence $3(x^2-3)(x^2-7)<0$. Thus 3<x²<7, so $-\sqrt7<x<-\sqrt3$ or $\sqrt3<x<\sqrt7$. The excluded poles lie outside these intervals. Multiplication by x²−1 without a sign split is unsafe.