Quotient curve analysis

Plan a quotient sketch

Use reciprocal-function features, partial fractions, quotient and logarithmic derivatives, and second-derivative signs. For a new quotient, record the domain, simplify on that domain, find intercepts and asymptotes, then use derivative signs for stationary points and concavity. A complete sketch combines all these facts

Keep the domain before cancelling

Record both denominator zeros as exclusions before cancelling

1. 1Cancel only on the original domain The reduced rule agrees with the original function only on its original domain \(\displaystyle f(x)=1+\frac{2}{x-1},\quad x\ne1,2\)
2. 2Keep the removed point At the cancelled input the limiting ordinate is finite, giving a hole \(\displaystyle \begin{gathered}x=2,\quad y=3\\\text{hole at }(2,3)\end{gathered}\)
3. 3Identify the pole The values diverge here, giving a vertical asymptote; a denominator zero is not automatically a pole \(\displaystyle x=1\)

Which point is removed in the plotted example?

Give the y-intercept ordinate for the original quotient

Reject a cancelled intercept

Keep the original exclusions when simplifying this quotient

1. 1Check the simplified zero The reduced rule has a zero at the cancelled input, but that input remains excluded \(\displaystyle g(x)=\frac{x-2}{x-1},\quad x\ne1,2\)
2. 2Hole, not intercept There are no horizontal-axis intercepts; the removed point is a hole \(\displaystyle \text{hole at }(2,0)\)
3. 3Read existence flags An Exists flag distinguishes an absent intercept from a real intercept whose coordinate is zero; a placeholder is never a graph point

Division reveals an oblique asymptote

Divide the numerator by the denominator and retain the excluded input

1. 1Divide exactly The remainder term tends to zero at both infinities \(\displaystyle \begin{gathered}f(x)=x-5+\frac4x\\f(x)-(x-5)=\frac4x\longrightarrow0\quad(x\to\pm\infty)\end{gathered}\)
2. 2Read the features The pole excludes a vertical-axis intercept; the two numerator zeros remain valid \(\displaystyle \begin{gathered}\text{Asymptotes: }x=0,\quad y=x-5\\\text{Intercepts: }(1,0),\ (4,0)\end{gathered}\)

Give the constant term of the oblique asymptote

Count distinct intercepts

The nonzero scale and positive integer parameter make the numerator roots distinct from the pole

1. 1Count distinct points The roots coincide only when the integer parameter is one, giving one intercept with multiplicity two \(\displaystyle \begin{gathered}x=h+1,\quad x=h+q^2\\q=1\ \Longrightarrow\ h+1=h+q^2\end{gathered}\)
2. 2Check the vertical-axis input Check the original denominator before substituting zero \(\displaystyle x=0\text{ is allowed}\iff h\ne0\)

Two poles and one-sided limits

Use division and the sign of the denominator on each side of a pole

1. 1Read the asymptotes and intercepts Neither numerator zero is excluded by the original denominator \(\displaystyle \begin{gathered}f(x)=2+\frac3{x^2-4}\\\text{Asymptotes: }x=-2,\ x=2,\ y=2\\f(0)=\frac54,\qquad x^2=\frac52\ \Longrightarrow\ x=\pm\sqrt{\frac52}\end{gathered}\)
2. 2Approach the negative pole The denominator is positive just to the left and negative just to the right \(\displaystyle \begin{gathered}\lim_{x\to-2^-}f(x)=+\infty\\\lim_{x\to-2^+}f(x)=-\infty\end{gathered}\)

Give the constant in the equation for the zeros

Degree comparison and the meaning of an asymptote

For a rational function, divide the polynomials first. A proper fraction tends to 0 at either infinity; equal degrees give a horizontal asymptote from the leading-coefficient ratio. A degree difference of one gives an oblique asymptote from division. Higher differences may give a polynomial asymptote. An asymptote describes limiting behaviour, not a line that a graph is forbidden to cross

Does the plotted function cross its horizontal asymptote?

Test symmetry with the domain

Check the domain as well as the function identity; removing only one of two opposite inputs can destroy symmetry

1. 1Require a symmetric domain Sign reversal must preserve membership of the domain \(\displaystyle x\in D\iff -x\in D\)
2. 2Compare function values Even functions preserve the value; odd functions reverse its sign \(\displaystyle \begin{gathered}\text{Even: }f(-x)=f(x)\\\text{Odd: }f(-x)=-f(x)\end{gathered}\)
3. 3Keep the quadratic denominator positive The nonzero parameter makes both examples defined on the whole real line \(\displaystyle \begin{gathered}a,q\in\mathbb R,\quad q\ne0,\quad x^2+q^2>0\\\frac{ax}{x^2+q^2}\text{ is odd on }\mathbb R\\\frac{a}{x^2+q^2}\text{ is even on }\mathbb R\end{gathered}\)

Stationary points of an odd quotient

Use the derivative signs to classify both stationary points

1. 1Differentiate and check signs The denominator is positive everywhere \(\displaystyle \begin{gathered}f'(x)=\frac{1-x^2}{(x^2+1)^2}\\f'(x)<0\text{ for }x<-1\text{ or }x>1\\f'(x)>0\text{ for }-1<x<1\end{gathered}\)
2. 2Classify the points A negative-to-positive change gives the minimum; a positive-to-negative change gives the maximum \(\displaystyle \begin{gathered}\text{Minimum: }(-1,-\tfrac12)\\\text{Maximum: }(1,\tfrac12)\\f'(0)=1\end{gathered}\)
3. 3Reverse the scale Multiplying by a negative constant reverses the derivative signs and both stationary natures \(\displaystyle (cf)'=cf',\qquad c<0\)

Give the maximum ordinate of the function

Inflections need a concavity change

Require a concavity change at a point in the domain

1. 1Find the second-derivative zeros The positive denominator leaves the numerator to determine the signs \(\displaystyle f''(x)=\frac{2x(x^2-3)}{(x^2+1)^3}\)
2. 2Check each interval The signs alternate across all three roots \(\displaystyle \begin{gathered}f''<0:\ (-\infty,-\sqrt3),\ (0,\sqrt3)\\f''>0:\ (-\sqrt3,0),\ (\sqrt3,\infty)\end{gathered}\)
3. 3Record the inflections All three inputs belong to the domain; the origin is not stationary \(\displaystyle \begin{gathered}(-\sqrt3,-\tfrac{\sqrt3}4),\quad(0,0),\quad(\sqrt3,\tfrac{\sqrt3}4)\\f'(0)=1\end{gathered}\)

Two branches with stationary points

Keep the two domain branches separate when checking derivative signs

1. 1Differentiate the divided form The first derivative vanishes at two allowed inputs \(\displaystyle \begin{gathered}f(x)=x-5+\frac4x\\f'(x)=1-\frac4{x^2},\qquad f''(x)=\frac8{x^3}\end{gathered}\)
2. 2Classify the stationary points The second derivative is negative at the negative input and positive at the positive input \(\displaystyle \begin{gathered}\text{Maximum: }(-2,-9)\\\text{Minimum: }(2,-1)\end{gathered}\)
3. 3Check concavity on the domain The pole is not a graph point, so the change across it is not an inflection \(\displaystyle \begin{gathered}f''<0\text{ for }x<0\\f''>0\text{ for }x>0\end{gathered}\)

How many inflections does this oblique-asymptote example have?

A quotient with two genuine inflections

The denominator is positive everywhere, so both second-derivative zeros are allowed

1. 1Differentiate twice The first derivative changes from positive to negative at zero \(\displaystyle \begin{gathered}f'(x)=\frac{-2x}{(x^2+3)^2}\\f''(x)=\frac{6(x^2-1)}{(x^2+3)^3}\\\text{Maximum: }(0,\tfrac13)\end{gathered}\)
2. 2Check the concavity changes The second derivative is positive outside the two roots and negative between them \(\displaystyle \begin{gathered}f''>0\text{ for }x<-1\text{ or }x>1\\f''<0\text{ for }-1<x<1\\\text{Inflections: }(-1,\tfrac14),\quad(1,\tfrac14)\end{gathered}\)

Give the common ordinate at the two given inflection inputs

Which interval is concave down in this figure?

An exponential quotient has two different ends

The exponential is positive, so the other factors determine the derivative signs

1. 1Find the stationary point The derivative changes from negative to positive at the sole stationary input \(\displaystyle \begin{gathered}f'(x)=\frac{e^{2x}(2x-1)}{x^2}\\\text{Minimum: }(\tfrac12,2e)\end{gathered}\)
2. 2Check concavity The second-derivative numerator is positive; the excluded zero prevents an inflection across the two branches \(\displaystyle \begin{gathered}f''(x)=\frac{e^{2x}[(2x-1)^2+1]}{x^3}\\f''<0\text{ for }x<0,\qquad f''>0\text{ for }x>0\end{gathered}\)
3. 3Check the axes The numerator never vanishes and zero is excluded, so neither axis has an intercept \(\displaystyle e^{2x}>0,\qquad 0\notin D_f\)

Give the exact stationary input of the function

Specify which infinity

State the direction of each limit; a horizontal asymptote at one end does not determine the other

1. 1Approach the pole The numerator approaches one while the denominator changes sign \(\displaystyle \begin{gathered}\lim_{x\to0^-}f(x)=-\infty\\\lim_{x\to0^+}f(x)=+\infty\\\text{Vertical asymptote: }x=0\end{gathered}\)
2. 2Compare the two ends The horizontal asymptote occurs only at the negative end; exponential growth dominates at the positive end \(\displaystyle \begin{gathered}\lim_{x\to-\infty}f(x)=0\\\lim_{x\to+\infty}f(x)=+\infty\\\text{Horizontal asymptote at the negative end: }y=0\end{gathered}\)

A logarithmic quotient has a restricted domain

The logarithm restricts the domain to positive inputs

1. 1Find the zero and differentiate The positive denominators make derivative signs depend on the bracketed expressions \(\displaystyle \begin{gathered}f(x)=0\iff x=\frac13\\f'(x)=\frac{2[1-\ln(3x)]}{x^2}\\f''(x)=\frac{2[2\ln(3x)-3]}{x^3}\end{gathered}\)
2. 2Classify the exact points The first derivative changes from positive to negative; the second changes from negative to positive \(\displaystyle \begin{gathered}\text{Maximum: }(\tfrac e3,\tfrac6e)\\\text{Inflection: }(\tfrac{e^{3/2}}3,\tfrac9{e^{3/2}})\end{gathered}\)
3. 3State the concavity intervals Both intervals lie within the original positive domain \(\displaystyle \begin{gathered}f''<0\text{ for }0<x<\frac{e^{3/2}}3\\f''>0\text{ for }x>\frac{e^{3/2}}3\end{gathered}\)

Write the maximum ordinate in the given form and enter the constant

Limits and intercepts for the logarithmic quotient

Use the positive domain when taking limits and checking intercepts

1. 1Check each end of the domain Logarithmic growth is slower than linear growth at the positive end \(\displaystyle \begin{gathered}\lim_{x\to0^+}f(x)=-\infty\\\lim_{x\to+\infty}f(x)=0\text{ from above}\end{gathered}\)
2. 2Allow an asymptote crossing The valid zero lies on the horizontal asymptote; the excluded origin prevents a vertical-axis intercept \(\displaystyle \begin{gathered}\text{Asymptotes: }x=0,\quad y=0\\f(\tfrac13)=0\end{gathered}\)

Explain removed points and symmetry, ungraded

Compare the quotient with its reduced rule and explain why the cancelled input is neither an intercept nor an inflection; then explain why the restricted square function is not even

1. 1Compare the quotient and reduced expression Keep the original domain when comparing the two formulas \(\displaystyle \begin{gathered}\begin{gathered}f(x)=\frac{(x-2)^2}{(x-2)(x-1)}\\x\ne1,2\end{gathered}\\\text{Reduced expression: }\frac{x-2}{x-1}\end{gathered}\)
2. 2Test symmetry of the restricted square Use both the expression and its stated domain in your ungraded explanation \(\displaystyle \begin{gathered}g(x)=x^2\\D_g=\mathbb R\setminus\{1\}\end{gathered}\)

Model: a function includes its domain

A simplified expression does not restore an input removed from the original domain

1. 1Distinguish the hole and pole The reduced rule has a finite limit at the cancelled input; the remaining denominator gives a pole \(\displaystyle \begin{gathered}f(x)=\frac{x-2}{x-1}\quad(x\ne1,2)\\\lim_{x\to2}f(x)=0\quad\Longrightarrow\quad\text{hole at }(2,0)\\\text{Pole: }x=1\end{gathered}\)
2. 2Test the restricted square domain The domain fails sign reversal, so the function is not even despite its expression \(\displaystyle \begin{gathered}g(x)=x^2,\quad D_g=\mathbb R\setminus\{1\}\\-1\in D_g,\qquad1\notin D_g\end{gathered}\)

Construct a complete quotient sketch, ungraded

Sketch the given function and give its domain, intercepts, both asymptotes, stationary points, monotonicity, concavity and one-sided limits at the excluded input; explain why there are no inflections and how the facts constrain both branches

Model: combine algebra and calculus

Combine the exact features below to constrain both branches; the complete drawing and explanation remain ungraded

1. 1Combine the algebraic features The excluded origin prevents a vertical-axis intercept \(\displaystyle \begin{gathered}\text{Intercepts: }(1,0),\ (4,0)\\\text{Asymptotes: }x=0,\quad y=x-5\end{gathered}\)
2. 2Combine stationary points and monotonicity Use the derivative signs separately on each branch \(\displaystyle \begin{gathered}\text{Maximum: }(-2,-9),\quad\text{minimum: }(2,-1)\\\text{Increasing: }(-\infty,-2),\ (2,\infty)\\\text{Decreasing: }(-2,0),\ (0,2)\end{gathered}\)
3. 3Combine concavity and one-sided limits There is no inflection because the only sign change across the branches occurs at an excluded input \(\displaystyle \begin{gathered}f''<0\ (x<0),\qquad f''>0\ (x>0)\\\lim_{x\to0^-}f(x)=-\infty,\qquad\lim_{x\to0^+}f(x)=+\infty\end{gathered}\)

Justify asymptotic behaviour, ungraded

Explain the asymptote crossing and justify the logarithmic limit using the given substitution and bound; distinguish exact reasoning from a finite plotting window

1. 1Compare the function and oblique line Explain why the graph can cross its oblique asymptote \(\displaystyle \begin{gathered}f(x)=x+\frac{x}{x^2+1}\\y=x\end{gathered}\)
2. 2Justify the growth comparison Use the substitution and bound to prove the stated limit \(\displaystyle \begin{gathered}\lim_{x\to+\infty}\frac{\ln x}{x}=0\\t=\ln x,\qquad e^t\ge\frac{t^2}2\quad(t>0)\end{gathered}\)

Model: compare the difference and use a bound

An exact limiting difference establishes an asymptote; a finite graph alone cannot prove a limit

1. 1Check the difference and crossing The difference tends to zero at both infinities and vanishes at the origin \(\displaystyle \begin{gathered}f(x)=x+\frac{x}{x^2+1}\\f(x)-x=\frac{x}{x^2+1}\longrightarrow0\quad(x\to\pm\infty)\\f(0)-0=0\quad\Longrightarrow\quad\text{crossing of }y=x\end{gathered}\)
2. 2Apply the growth bound The upper bound tends to zero, so the squeeze theorem gives the limit \(\displaystyle \begin{gathered}t=\ln x>0,\quad x=e^t\\0<\frac{\ln x}{x}=\frac t{e^t}\le\frac2t\longrightarrow0\quad(t\to+\infty)\end{gathered}\)
3. 3Justify the bound The positive third derivative and zero initial values make each lower derivative positive for positive inputs \(\displaystyle \begin{gathered}H(t)=e^t-1-t-\frac{t^2}2\\H'''(t)=e^t>0,\quad H''(0)=H'(0)=H(0)=0\\H(t)>0\ (t>0)\quad\Longrightarrow\quad e^t>1+t+\frac{t^2}2>\frac{t^2}2\end{gathered}\)

Investigate plotting restrictions with technology, ungraded

Plot these two functions in two windows and record whether the software marks the removed point or joins branches across a pole; compare the picture with the exact domains and one-sided limits, add the hole and asymptotes if needed, and keep your graph and explanation

1. 1Original quotient Retain both exclusions before simplifying \(\displaystyle \begin{gathered}f(x)=\frac{(x-2)(x+1)}{(x-2)(x-1)}\\D_f=\mathbb R\setminus\{1,2\}\end{gathered}\)
2. 2Reduced rule on its natural domain This second function includes the cancelled input, so compare the domains as well as the plotted curves \(\displaystyle \begin{gathered}g(x)=\frac{x+1}{x-1}=1+\frac2{x-1}\\D_g=\mathbb R\setminus\{1\}\end{gathered}\)

Model: distinguish a picture from the function

The two rules agree on the original domain, but plotting software can simplify away a hole or connect coarse samples across a pole; exact domains and limits determine the graph

1. 1Compare the domains The original function has a hole at the input included by the natural reduced rule \(\displaystyle \begin{gathered}\begin{gathered}f(x)=\frac{(x-2)(x+1)}{(x-2)(x-1)}\\D_f=\mathbb R\setminus\{1,2\}\end{gathered}\\\begin{gathered}g(x)=\frac{x+1}{x-1}=1+\frac2{x-1}\\D_g=\mathbb R\setminus\{1\}\end{gathered}\\g(2)=3\quad\Longrightarrow\quad\text{original hole at }(2,3)\end{gathered}\)
2. 2Keep the branches separate A plotting-window edge is not a domain endpoint; this model is not evidence that the learner operated the software \(\displaystyle \begin{gathered}\lim_{x\to1^-}f(x)=-\infty\\\lim_{x\to1^+}f(x)=+\infty\\\text{Asymptotes: }x=1,\quad y=1\end{gathered}\)

Exact features and complete sketches

Find exact features of rational, exponential and logarithmic quotients