Unit 2 - Functions
Get Ready
Using the given graph below, adjust the slider to see the following graphs:
a) \(y=\frac{1}{x}\)
b) \(y=\frac{2}{x}\)
c) \(y=\frac{4}{x}\)
d) \(y=-\frac{1}{x}\)
e) \(y=-\frac{2}{x}\)
f) \(y=-\frac{4}{x}\)
Describe the effects of varying \(k\) in the function \(f\left(x\right)=\frac{k}{x}\).
Notes
The reciprocal function is a function that can be written in the form $$f\left(x\right)=\frac{k}{x}$$The graph of a reciprocal function is called a hyperbola and is a conic section, much like a parabola. It has a distinct shape with two branches.
The graph has two asymptotes:
A vertical asymptote at \(x=0\)
A horizontal asymptote at \(y=0\)
The natural domain of a reciprocal function is \(\left\{x\ne 0\right\}\).
The range of a reciprocal function is \(\left\{y\ne 0\right\}\).
All reciprocal functions of the form \(f\left(x\right)=\frac{k}{x}\) pass through the points$$\left(1,k\right)\quad,\quad\left(k,1\right)\quad,\quad\left(-1,-k\right)\quad,\quad\left(-k,-1\right)$$
***Important note***
The function \(f\left(x\right)=\frac{1}{2x}\) is a reciprocal function as it can be written as$$f\left(x\right)=\frac{1}{2x}=\frac{1}{2}\times\frac{1}{x}=\frac{1\over 2}{x}$$
Examples and Your Turns
Example
Determine the equation of the reciprocal function below.
Your Turn
Determine the equation of the reciprocal function below.
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This is a reciprocal function so the general form is$$y=\frac{k}{x}$$Substituting the known point \(\left(-5,-3\right)\) we get$$-3=\frac{k}{-5}$$Solving for \(k\) we find$$k=15$$Thus the equation is$$y=\frac{15}{x}$$
Your Turn
Determine the equation of the reciprocal function below.
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This is a reciprocal function so the general form is$$y=\frac{k}{x}$$Substituting the known point \(\left(-9,4\right)\) we get$$4=\frac{k}{-9}$$Solving for \(k\) we find$$k=-36$$Thus the equation is$$y=-\frac{36}{x}$$
Your Turn
Determine the equation of the reciprocal function below.
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This is a reciprocal function so the general form is$$y=\frac{k}{x}$$Substituting the known point \(\left(-2,0.75\right)\) we get$$0.75=\frac{k}{-2}$$Solving for \(k\) we find$$k=-1.5$$Thus the equation is$$y=-\frac{1.5}{x}=-\frac{3}{2x}$$
Notes
A rational function is a function that can be written in the form $$\frac{g\left(x\right)}{h\left(x\right)},\quad h\left(x\right)\ne 0$$Reciprocal functions are special cases of rational functions.
In this course, the functions \(g\left(x\right)\) and \(h\left(x\right)\) will always be constant, linear or quadratic.
Linear rational functions can be written in one of two ways:
$$y=\frac{ax+b}{cx+d}$$
$$y=a+\frac{b}{cx+d}$$
Note that in the same function can be written in these two forms, but with different values of \(a\) and \(b\).
Investigation
Below is the graph of \(y=a+\frac{b}{cx+d}\). By using the sliders to change the values of the constants, determine:
the equation of the vertical asymptote
the equation of the horizontal asymptote
the domain and range
why \(b\ne 0\)
why \(c\ne 0\)
We will now do the same for the graph of \(y=\frac{ax+b}{cx+d}\). By using the sliders to change the values of the constants, determine:
the equation of the vertical asymptote
the equation of the horizontal asymptote
the domain and range
why \(b\ne 0\)
why \(c\ne 0\)
Notes
For the function $$f\left(x\right)=a+\frac{b}{cx+d}$$we know the following:
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The vertical asymptote of $$f\left(x\right)=a+\frac{b}{cx+d}$$ is given by $$x=-\frac{d}{c}$$This is because it occurs when the denominator is equal to \(0\), so \(cx+d=0\).
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The horizontal asymptote of $$f\left(x\right)=a+\frac{b}{cx+d}$$ is given by $$y=a$$This is because we are looking for what happens as \(x\to\infty\). As \(x\to\infty\),we see that \(cx+d\to\infty\) too, but this is the denominator of the fraction, which means \(\frac{b}{cx+d}\to\infty\). This leaves us with \(y=a\).
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The domain of $$f\left(x\right)=a+\frac{b}{cx+d}$$is all values of \(x\) except the vertical asymptote, so $$\left\{x:x\ne -\frac{d}{c}\right\}$$
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The range of $$f\left(x\right)=a+\frac{b}{cx+d}$$is all values of \(y\) except the horizontal asymptote, so $$\left\{y:y\ne a\right\}$$
For the function $$f\left(x\right)=\frac{ax+b}{cx+d}$$we know the following:
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The vertical asymptote of $$f\left(x\right)=\frac{ax+b}{cx+d}$$ is given by $$x=-\frac{d}{c}$$This is because it occurs when the denominator is equal to \(0\), so \(cx+d=0\).
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The horizontal asymptote of $$f\left(x\right)=\frac{ax+b}{cx+d}$$ is given by $$y=\frac{a}{c}$$This is because we are looking for what happens as \(x\to\infty\). As \(x\to\infty\),we see that \(ax+b\to ax\) and \(cx+d\to cx\). This means \(f\left(x\right)\to\frac{ax}{cx}=\frac{a}{c}.
Another way to think about this is that when \(x\) is REALLY large, the \(+b\) is tiny and irrelevant compared to the \(ax\). So it basically becomes \(ax\). A similar argument works in the denominator.
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The domain of $$f\left(x\right)=\frac{ax+b}{cx+d}$$is all values of \(x\) except the vertical asymptote, so $$\left\{x:x\ne -\frac{d}{c}\right\}$$
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The range of $$f\left(x\right)=\frac{ax+b}{cx+d}$$is all values of \(y\) except the horizontal asymptote, so $$\left\{y:y\ne \frac{a}{c}\right\}$$
As with all functions, it is not really important to remember these specific rules, but rather the general method / principles that work for ANY function.
To find
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Set \(x=0\) and solve for the \(y\) value.
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Set \(y=0\) and solve for the \(x\) value(s).
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Set the denominator equal to \(0\) and solve for \(x\). Write it as an equation of the form \(x=\).
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Consider what happens as \(x\to\infty\), which leaves a \(y\) value. Write it as an equation of the form \(y=\).
Examples and Your Turns
Example
Consider the function$$y=\frac{6}{x-2}+4, x\ne 2$$
a) Find the equations of the asymptotes of the function.
b) Find the axes intercepts.
c) Sketch the graph of the function, clearly indicating what you found in (a) and (b).
Your Turn
Consider the function$$y=\frac{3}{x+1}-2, x\ne -1$$
a) Find the equations of the asymptotes of the function.
b) Find the axes intercepts.
c) Sketch the graph of the function, clearly indicating what you found in (a) and (b).
Example
Sketch the graph of the function$$f\left(x\right)=\frac{3x-8}{2x+6}, x\ne -3$$clearly indicating the equations of the asymptotes and any axes intercepts.
Your Turn
Sketch the graph of the function$$f\left(x\right)=\frac{4x+1}{3x-1}, x\ne \frac{1}{3}$$clearly indicating the equations of the asymptotes and any axes intercepts.
Notes (HL)
For any rational function of the form \(f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}\), the vertical asymptote(s) will occur when \(h\left(x\right)=0\).
Horizontal asymptotes occur if as \(x\to\infty\) then \(f\left(x\right)\to k\) for some \(k\in\mathbb{R}\).
Investigation (HL)
For each of the following functions, reveal the graph and use it to write down the equation of any vertical or horizontal asymptotes.
Try and generalise what you have found.
For a function of the form \(y=\frac{ax+b}{cx^{2}+dx+e}, c\ne 0\), the horizontal asymptote is…
We can determine any vertical asymptotes by…
Can you explain why these occur, thinking about limits.
Examples and Your Turns (HL)
Example
Consider the function$$y=\frac{2x-6}{x^{2}-3x-4}$$
a) Find the equations of all asymptotes to the function.
b) Find the coordinates of all axes intercepts.
c) Sketch the graph of the function, clearly indicating what you found in (a) and (b).
Your Turn
Consider the function$$y=\frac{3x+2}{x^{2}-5x+6}$$
a) Find the equations of all asymptotes to the function.
b) Find the coordinates of all axes intercepts.
c) Sketch the graph of the function, clearly indicating what you found in (a) and (b).
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
Why do vertical asymptotes occur when the denominator is equal to \(0\)?
How can you determine if a rational function crosses its horizontal asymptote?
Common Mistakes / Misconceptions
The biggest difficulty with this skill is determining the horizontal asymptote. Remember to think about what happens as \(x\to\infty\).
Thinking the horizontal asymptote is always \(y=0\) which is not true for a linear over a linear.
Poor quality sketches where the graph bends away from the asymptote instead of approaching it, or with roots in wrong locations.
Connecting This to Other Skills
Strong algebraic skills are necessary for this skill, such as solving quadratic equations (PK3) and working with algebraic fractions (PK5).
We will be directly building on this idea in Further Rational Functions (2.25), where we look at more types.
Polynomials (2.19) are the foundation blocks of more complicated rational functions.
It is a common question to have to solve an Inequality (2.24) involving rational functions.
In calculus, the Quotient Rule (4.10) is all about differentiating rational functions.
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?