A rational number is a number in the form of a fraction, and similarly a rational function is a function in the form of a fraction. So....
A rational function is a function of the form $y=\frac{P\left(x\right)}{Q\left(x\right)},Q\left(x\right)\ne0$y=P(x)Q(x),Q(x)≠0, where $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are polynomials.
So functions like $y=\frac{x}{x+1}$y=xx+1, $y=\frac{x^2-9}{2x-1}$y=x2−92x−1, $y=5x^3-1=\frac{5x^3-1}{1}$y=5x3−1=5x3−11 and $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6 can all be considered rational functions.
Any fraction, written as $\frac{m}{n}$mn, where $m$m and $n$n are integers (and $n$n is always nonzero), has the value zero if, and only if, $m=0$m=0. The same idea applies to rational functions. The function values becomes zero when the numerator vanishes (in other words when $P\left(x\right)=0$P(x)=0). The complication with rational functions is that sometimes the denominator vanishes at the same time.
For example, $y=\frac{x^2-5x+6}{x}$y=x2−5x+6x becomes zero when $x^2-5x+6=0$x2−5x+6=0. This means that all of the function's $x$x intercepts are found by solving this quadratic equation:
$x^2-5x+6$x2−5x+6 | $=$= | $0$0 |
$\left(x-2\right)\left(x-3\right)$(x−2)(x−3) | $=$= | $0$0 |
$\therefore$∴ $x$x | $=$= | $2,3$2,3 |
So because the values $x=2$x=2 and $x=3$x=3 don't send the denominator to $0$0 as well, we can accept these as the two $x$x intercepts of this function. We need to make the check that the denominator isn't being made equal to 0, because if the denominator is 0, then the function is undefined at that point.
An asymptote is a line (or curve) that approaches a given curve arbitrarily closely. The curve approaches it, but never reaches it.
To appreciate this, consider the simple rational function given by $y=\frac{x-1}{x}$y=x−1x and imagine that the positive $x$x values becomes very large. The numerator will become large as well, but will always be one less than the denominator, and this means that the function values will become close to $1$1 but never quite reach $1$1. Hence the line $y=1$y=1 becomes a horizontal asymptote.
In addition to this we find that rational functions are rarely continuous over their natural domains. There are often points of discontinuity where the function simply ceases to exist.
Take for example the rational function given by $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6. If the denominator becomes $0$0, the function becomes undefined, and because we know that $x=2$x=2 and $x=3$x=3 are roots of the equation $x^2-5x+6=0$x2−5x+6=0 (see above), then the curve must break into three sections across the vertical lines given by $x=2$x=2 and $x=3$x=3.
We need to examine the behaviour of the function as the $x$x values approach each of these lines.
For example we know that at $x=3$x=3, the denominator becomes $0$0, so at $x=3.000001$x=3.000001, which is slightly more than $x=3$x=3, the denominator must be extremely small (in fact it becomes the number $0.000001000001$0.000001000001).
The function value is thus given by the constant $12$12 (in the numerator) divided by this extremely small number. This results in an extremely large number (almost $24,000,000$24,000,000).
What this means is that as $x$x approaches $3$3 from the right side of the line $x=3$x=3, the curve bends upward toward infinity. It's as if the curve, upon nearing the line $x=3$x=3, tries to find a way around it. The line $x=3$x=3 becomes a vertical asymptote.
Here is how the right hand section of the function $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6 looks. Note how the curve bends upward as it approaches the vertical asymptote $x=3$x=3:
The same type of behaviour occurs on the other side of the line $x=3$x=3, except that with this particular function, the curve approaches the line $x=3$x=3 from the left bending downwards, as if it's trying to find a way around it in the other direction.
With the asymptote $x=2$x=2, we see the curve behaving in a similar way. On both sides it approaches and bends in opposite directions toward the line $x=2$x=2.
Here is a complete sketch of the curve with a very much compressed scale on the $y$y axis. Note the two vertical asymptotes as well as the horizontal asymptote at $y=0$y=0 (the $x$x axis).
Consider the function given by $y=\frac{x^2+4}{x}$y=x2+4x.
Clearly we have $x\ne0$x≠0, and so $x=0$x=0 is a vertical asymptote of the function.
Further, as $x$x becomes large, the function approaches the line $y=x$y=x. The easiest way to see this is to put the function into a different form by carrying out the division by $x$x. Thus:
$y$y | $=$= | $\frac{x^2+4}{x}$x2+4x |
$=$= | $x+\frac{4}{x}$x+4x | |
As x increases indefinitely (that is as $x\rightarrow\infty$x→∞ ), the fractional part $\frac{4}{x}$4x becomes very small (in other words $\frac{4}{x}\rightarrow0$4x→0), so that the curve approaches the line $y=x$y=x. It does so from above because $x+\frac{4}{x}$x+4x is always greater than $x$x for all $x>0$x>0.
As $x\rightarrow-\infty$x→−∞ then $\frac{4}{x}\rightarrow0$4x→0 (always less than zero) and this means that the curve $y=x+\frac{4}{x}$y=x+4x approaches the line $y=x$y=x from below.
We also know that there are no $x$x intercepts, because setting $y=0$y=0 implies that the numerator function $x^2+4$x2+4 has to be $0$0, but this is impossible because both $x^2+4$x2+4 is always positive (check this).
Putting all this together, the graph of the function, shown here, makes sense.
Consider the following graph of a hyperbola.
State the equation of the vertical asymptote.
State the equation of the horizontal or oblique asymptote.
Consider the function $f\left(x\right)=\frac{7x-5}{4x+3}$f(x)=7x−54x+3.
State the equations of any vertical asymptotes.
State the equations of any horizontal or oblique asymptotes.
Consider the function $f\left(x\right)=\frac{x^2+8x+15}{3x^2-17x+10}$f(x)=x2+8x+153x2−17x+10.
State the equations of any vertical asymptotes.
State the equations of any horizontal or oblique asymptotes.
(Your answer should be given in terms of $y$y ).
A rational number is a number in the form of a fraction, and similarly a rational function is a function in the form of a fraction. So....
A rational function is a function of the form $$, where $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are polynomials.
So functions like $y=\frac{x}{x+1}$y=xx+1, $y=\frac{x^2-9}{2x-1}$y=x2−92x−1, $y=5x^3-1=\frac{5x^3-1}{1}$y=5x3−1=5x3−11 and $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6 can all be considered rational functions.
Any fraction, written as $\frac{m}{n}$mn, where $m$m and $n$n are integers (and $n$n is always nonzero), has the value zero if, and only if, $m=0$m=0. The same idea applies to rational functions. The function values becomes zero when the numerator vanishes (in other words when $P\left(x\right)=0$P(x)=0). The complication with rational functions is that sometimes the denominator vanishes at the same time.
For example, $y=\frac{x^2-5x+6}{x}$y=x2−5x+6x becomes zero when $x^2-5x+6=0$x2−5x+6=0. This means that all of the function's $x$x intercepts are found by solving this quadratic equation:
$x^2-5x+6$x2−5x+6 | $=$= | $0$0 |
$\left(x-2\right)\left(x-3\right)$(x−2)(x−3) | $=$= | $0$0 |
$\therefore$∴ $x$x | $=$= | $2,3$2,3 |
So because the values $x=2$x=2 and $x=3$x=3 don't send the denominator to $0$0 as well, we can accept these as the two $x$x intercepts of this function. We need to make the check that the denominator isn't being made equal to 0, because if the denominator is 0, then the function is undefined at that point.
An asymptote is a line (or curve) that approaches a given curve arbitrarily closely. The curve approaches it, but never reaches it.
To appreciate this, consider the simple rational function given by $y=\frac{x-1}{x}$y=x−1x and imagine that the positive $x$x values becomes very large. The numerator will become large as well, but will always be one less than the denominator, and this means that the function values will become close to $1$1 but never quite reach $1$1. Hence the line $y=1$y=1 becomes a horizontal asymptote.
In addition to this we find that rational functions are rarely continuous over their natural domains. There are often points of discontinuity where the function simply ceases to exist.
Take for example the rational function given by $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6. If the denominator becomes $0$0, the function becomes undefined, and because we know that $x=2$x=2 and $x=3$x=3 are roots of the equation $x^2-5x+6=0$x2−5x+6=0 (see above), then the curve must break into three sections across the vertical lines given by $x=2$x=2 and $x=3$x=3.
We need to examine the behaviour of the function as the $x$x values approach each of these lines.
For example we know that at $x=3$x=3, the denominator becomes $0$0, so at $x=3.000001$x=3.000001, which is slightly more than $x=3$x=3, the denominator must be extremely small (in fact it becomes the number $0.000001000001$0.000001000001).
The function value is thus given by the constant $12$12 (in the numerator) divided by this extremely small number. This results in an extremely large number (almost $24,000,000$24,000,000).
What this means is that as $x$x approaches $3$3 from the right side of the line $x=3$x=3, the curve bends upward toward infinity. It's as if the curve, upon nearing the line $x=3$x=3, tries to find a way around it. The line $x=3$x=3 becomes a vertical asymptote.
Here is how the right hand section of the function $y=\frac{24}{x^2-5x+6}$y=24x2−5x+6 looks. Note how the curve bends upward as it approaches the vertical asymptote $x=3$x=3:
The same type of behaviour occurs on the other side of the line $x=3$x=3, except that with this particular function, the curve approaches the line $x=3$x=3 from the left bending downwards, as if it's trying to find a way around it in the other direction.
With the asymptote $x=2$x=2, we see the curve behaving in a similar way. On both sides it approaches and bends in opposite directions toward the line $x=2$x=2.
Here is a complete sketch of the curve with a very much compressed scale on the $y$y axis. Note the two vertical asymptotes as well as the horizontal asymptote at $y=0$y=0 (the $x$x axis).
Consider the function given by $y=\frac{x^2+4}{x}$y=x2+4x.
Clearly we have $x\ne0$x≠0, and so $x=0$x=0 is a vertical asymptote of the function.
Further, as $x$x becomes large, the function approaches the line $y=x$y=x. The easiest way to see this is to put the function into a different form by carrying out the division by $x$x. Thus:
$y$y | $=$= | $\frac{x^2+4}{x}$x2+4x |
$=$= | $x+\frac{4}{x}$x+4x | |
As x increases indefinitely (that is as $x\rightarrow\infty$x→∞ ), the fractional part $\frac{4}{x}$4x becomes very small (in other words $\frac{4}{x}\rightarrow0$4x→0), so that the curve approaches the line $y=x$y=x. It does so from above because $x+\frac{4}{x}$x+4x is always greater than $x$x for all $x>0$x>0.
As $x\rightarrow-\infty$x→−∞ then $\frac{4}{x}\rightarrow0$4x→0 (always less than zero) and this means that the curve $y=x+\frac{4}{x}$y=x+4x approaches the line $y=x$y=x from below.
We also know that there are no $x$x intercepts, because setting $y=0$y=0 implies that the numerator function $x^2+4$x2+4 has to be $0$0, but this is impossible because both $x^2+4$x2+4 is always positive (check this).
Putting all this together, the graph of the function, shown here, makes sense.
Consider the following graph of a hyperbola.
State the equation of the vertical asymptote.
State the equation of the horizontal or oblique asymptote.
Consider the function $f\left(x\right)=\frac{7x-5}{4x+3}$f(x)=7x−54x+3.
State the equations of any vertical asymptotes.
State the equations of any horizontal or oblique asymptotes.
Consider the function $f\left(x\right)=\frac{x^2+8x+15}{3x^2-17x+10}$f(x)=x2+8x+153x2−17x+10.
State the equations of any vertical asymptotes.
State the equations of any horizontal or oblique asymptotes.
(Your answer should be given in terms of $y$y ).
Form and use trigonometric, polynomial, and other non-linear equations
Apply trigonometric methods in solving problems