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Read a rational function off its factors — asymptotes, holes, intercepts and end behaviour — and transform any graph by working inside and outside the function.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you take two families of question that used to be answered by plotting points and answer them by reading a formula instead. First, rational functions: factoring the top and the bottom tells you where the graph has a wall, where it has a single missing point, where it crosses the axes and what line it settles onto far away. Then transformations, gathered into one rule: changes inside the function act horizontally and in reverse, changes outside act vertically and as they read, and even and odd are that rule asked about $-x$.
You can factor a quadratic and cancel a common factor from a fraction, and you know that dividing by zero is not allowed. You also know the degree of a polynomial: the highest power of $x$ in it. This lesson asks one new question of those old skills — not what is the value here? but what does the graph do near the places where there is no value, and far away where $x$ is enormous?
Rational function: a ratio of two polynomials, $f(x) = \frac{P(x)}{Q(x)}$.
Vertical asymptote: a line $x = c$ the graph runs alongside without ever touching, because $Q(c) = 0$ and the fraction blows up.
Hole: a single missing point, left when a factor cancels from top and bottom.
Horizontal asymptote: a line $y = L$ the graph settles onto as $x$ runs to $\pm\infty$.
Slant asymptote: the same idea, but the line is tilted.
End behaviour: what the graph does far to the left and far to the right.
Factor the top and the bottom first — everything else is read off the factors. A zero of the denominator that does not cancel gives a vertical asymptote: $\frac{2x + 1}{x - 3}$ has $x = 3$, and $\frac{x + 1}{x^2 - 9}$ has $x = -3$ and $x = 3$. A factor that does cancel leaves a hole: $\frac{x^2 - 4}{x - 2}$ is the line $y = x + 2$ with the point $(2, 4)$ punched out. Zeros of the numerator that survive are $x$-intercepts, so $\frac{x - 5}{x + 2}$ crosses at $x = 5$, and $f(0)$ is the $y$-intercept. End behaviour comes from the degrees: equal degrees give $y = $ the ratio of the leading coefficients, so $\frac{2x + 1}{x - 3}$ flattens onto $y = 2$; a smaller top gives $y = 0$; a top exactly one degree higher gives a slant asymptote, found by dividing, so $\frac{x^2 + 1}{x - 1}$ follows $y = x + 1$.
Another way: picture
The graph of $\frac{2x + 1}{x - 3}$: a dashed vertical line at $x = 3$ and a dashed horizontal line at $y = 2$, with two branches in opposite corners, each hugging both dashed lines as it runs away.
Another way: story
Near a denominator zero the bottom is tiny, so the fraction is huge — that is the wall. Far out, the leading terms are the only ones that matter, so the function copies their ratio — that is the shelf it settles on.
"Every denominator zero is an asymptote." Only the ones that survive the cancelling. In $\frac{x^2 - 4}{x - 2}$ the denominator is zero at $2$, but the factor cancels and the graph has a hole there, not a wall.
"A graph can never cross its horizontal asymptote." It often does, in the middle. The asymptote is a promise about the far left and the far right, not a fence.
"Bigger degree on top means a horizontal asymptote higher up." It means there is no horizontal asymptote at all: one degree more gives a slant asymptote, two or more and the graph runs away entirely.
Factor the bottom: $x^2 - 9 = (x - 3)(x + 3)$. Neither factor matches $x + 1$.
Factor first, then look for cancellation.
Nothing cancels, so both zeros are vertical asymptotes: $x = -3$ and $x = 3$.
Top degree $1$ is less than bottom degree $2$, so the horizontal asymptote is $y = 0$.
A smaller top wins the race to zero.
The top degree is one more than the bottom, so divide: $x^2 + 1 = (x - 1)(x + 1) + 2$.
Long division, or spotting the product.
So $f(x) = x + 1 + \frac{2}{x - 1}$, and the remainder term dies away: the graph follows $y = x + 1$.
Factor the top: $3x - 6 = 3(x - 2)$, and the factor $x - 2$ cancels.
So the graph is the horizontal line $y = 3$ with a hole at $(2, 3)$ — no asymptote at all.
$f(x) = \dfrac{4x - 5}{x - 6}$. Give the vertical asymptote and the horizontal asymptote.
x = v and y = h
$f(x) = \dfrac{x^2 - 25}{x - 5}$. What happens at $x = 5$?
How does $f(x) = \dfrac{x^2 + 1}{x - 7}$ behave when $x$ is very large?
Where does $f(x) = \dfrac{6x - 10}{x + 2}$ cross the $y$-axis?
Answer:
Brine runs into a tank of pure water. After $t$ minutes the salt concentration is $C(t) = \dfrac{7t}{t + 31}$ grams per litre. Left running all day, what does the concentration settle at?
You have shifted a parabola before, and you know the graphs of $y = x^2$, $y = \sqrt{x}$ and $y = |x|$ by heart. You also know function notation: $f(x)$ names the output for the input $x$. That notation is what makes this lesson short — every transformation is a change either to what goes into $f$ or to what comes out of it, and those two behave differently.
Translation (shift): sliding a graph without changing its shape.
Vertical stretch by $a$: multiplying every output by $a$, so the graph is pulled away from the $x$-axis.
Horizontal compression by $k$: replacing $x$ by $kx$, so the graph is squeezed towards the $y$-axis.
Reflection: a mirror image — in the $x$-axis for $-f(x)$, in the $y$-axis for $f(-x)$.
Even function: $f(-x) = f(x)$, mirror-symmetric in the $y$-axis.
Odd function: $f(-x) = -f(x)$, unchanged by a half-turn about the origin.
Every transformation is one of two kinds. Inside the function is the input's world, and it behaves backwards: $f(x - h)$ shifts right by $h$, $f(x + h)$ shifts left, $f(kx)$ compresses horizontally by a factor of $k$, and $f(-x)$ reflects in the $y$-axis. Outside is the output's world, and it behaves as it reads: $f(x) + k$ shifts up by $k$, $a\,f(x)$ stretches vertically by $a$, and $-f(x)$ reflects in the $x$-axis. So $y = x^2$ moved $3$ right and $2$ up is $(x - 3)^2 + 2$; $y = \sqrt{x}$ stretched by $3$ and reflected is $-3\sqrt{x}$; $y = |x|$ moved $2$ left and reflected is $-|x + 2|$. When several act at once, follow the order of operations on $x$: for $2f(x - 1) + 5$, shift right, then stretch, then shift up. Symmetry is the same test applied to $-x$: even functions satisfy $f(-x) = f(x)$, odd ones $f(-x) = -f(x)$, so an odd $f$ with $f(3) = 7$ must have $f(-3) = -7$.
Another way: picture
The parabola $y = x^2$ drawn faintly, and $y = (x - 3)^2 + 2$ drawn solid beside it, with the vertex marked at $(0,0)$ and again at $(3, 2)$, an arrow three units right and an arrow two units up between them.
Another way: story
Inside the brackets you are talking to the input, and the input argues: ask for $x - 3$ and the graph goes the other way. Outside you are talking to the output, and the output simply obeys.
"$f(x - 3)$ moves the graph left, because of the minus." It moves it right. To get the old output you now have to feed in an $x$ that is $3$ bigger, so every point slides right by $3$.
"$f(2x)$ stretches the graph horizontally." It compresses it by a factor of $2$: the input reaches every value twice as fast, so the picture is half as wide.
"Order does not matter." For $2f(x) + 1$ it does. Stretch first, then shift up; shifting first and stretching afterwards would multiply the $+1$ as well and land you two units too high.
Left is an input change, and it goes in backwards: $y = |x + 2|$.
Inside the bars, reversed.
Reflecting in the $x$-axis negates the output: $y = -|x + 2|$.
Outside, as it reads.
Work out $f(-x) = (-x)^3 - 4(-x) = -x^3 + 4x$.
Substitute $-x$ everywhere.
That is exactly $-(x^3 - 4x) = -f(x)$, so $f$ is odd: the graph is unchanged by a half-turn about the origin.
Inside: $x + 1$ shifts the graph of $\sqrt{x}$ one unit left.
Outside: $\times 2$ stretches it vertically by $2$, and the minus reflects it in the $x$-axis.
Write the function whose graph is $y = x^2$ moved $3$ to the right and $7$ up.
Answer:
Which formula has the graph of $y = \sqrt{x}$ stretched vertically by $3$ and then reflected in the $x$-axis?
$f$ is an odd function and $f(7) = 8$. What is $f(-7)$?
Answer:
Put the steps in the order that turns the graph of $y = f(x)$ into $y = 2f(x - 2) + 4$.
Number the steps in order (write the number in the box):
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = \dfrac{6x - 8}{x - 4}$. Give the vertical asymptote and the horizontal asymptote.
x = v and y = h
Write the function whose graph is $y = x^2$ moved $6$ to the right and $6$ up.
Answer:
You can describe a rational function from its factors and transform any graph. From memory: why does $\frac{x^2 - 4}{x - 2}$ have a hole rather than an asymptote at $x = 2$, and which way does the graph of $f(x - 5)$ move?
8. Your turn: describe $\frac{3x - 6}{x - 2}$, step 2
20. Your turn: describe $y = -2\sqrt{x + 1}$, step 2