Download Equations Determining Lines

Math 31A Precalculus Review
Austin Christian
Winter 20161
This document briefly covers a collection of precalculus topics that you should feel quite
comfortable with coming into the term. If any of these topics are new or challenging to you,
you should seek out help early in order to catch up. Khan Academy is a great place to brush
up on your precalculus. There are a handful of examples in these notes; solutions can be
found at the end.
Equations Determining Lines
The central idea of differential calculus is to (locally) treat curves as if they were lines; one
reason we do this is that equations determining lines are generally easier to manipulate than
equations determining curves. In particular, before we can start doing calculus we need to
be proficient in computing slopes of lines and in finding equations for lines having various
properties. Recall that given two points (x1 , y1 ) and (x2 , y2 ) in R2 , the slope of the line
between these points measures the ratio of vertical change to horizontal change as we move
from one point to the other:
slope between x1 and x2 =
y2 − y1
vertical difference between two points
=
,
horizontal difference between two points
x2 − x1
provided x1 6= x2 . If x1 = x2 , we don’t have any horizontal change, so we don’t define a slope
between these points. Notice that if we draw a line passing through (x1 , y1 ) and (x2 , y2 ) and
choose a third point (x, y), the slope between this third point and (x1 , y1 ) will be the same
as the slope between (x1 , y1 ) and (x2 , y2 ), so
y2 − y1
y − y1
=
.
x − x1
x2 − x1
We can rearrange this to write
y=
y2 − y1
(x − x1 ) + y1 .
x2 − x1
(1)
So every point lying on the line that passes through (x1 , y1 ) and (x2 , y2 ) will satisfy equation
(1), and we say that (1) gives us an equation for the line passing through (x1 , y1 ) and (x2 , y2 ).
Examples.
1. Find the equation of the line passing through (1, 4) with slope 2. Where does this line
intersect the line x = 5?
2. Find the equation of the line passing through (−1, 1) and (3, 9). Where does this line
intersect the line y = 13?
1
Last edited: December 18, 2015
1
Rational Functions
In calculus we must be able to manipulate many different types of functions and expressions. Here we’ll review rational functions, one of the most common types of functions
for students to forget about. A rational function is a function of the form
f (x) =
an xn + an−1 xn−1 + · · · + a1 x + a0
p(x)
=
,
q(x)
bm xm + bm−1 xm−1 + · · · + b1 x + b0
where p and q are polynomial functions of degree n and m, respectively. Like polynomials,
rational functions are continuous2 wherever they are defined, but unlike polynomials, rational functions are not defined for all real numbers. Namely, if q(x) = 0 for some x ∈ R, then
f (x) is not defined, since the definition of f would cause us to divide by zero.
Example.
3. For what values is the rational function f (x) =
(6x − 12)(x2 − 4x + 3)
defined?
(2x − 6)(x2 − x − 20)
Graphically, a rational function will have one of two responses to a point which is not in
its domain: the function may just skip over the x-value for which it is not defined, as you
see for x = 3 in the figure below, or the function may “blow up”, as you see at x = −4 and
x = 5.
The function used in this graph is the same function used in Example 3. The reason f
behaves so much more nicely near x = 3 has to do with the factors of its numerator and
denominator:
3(x − 2)(x − 3)(x − 1)
f (x) =
.
(x − 3)(x − 5)(x + 4)
Because both the numerator and denominator of f have a factor of x − 3, we have
f (x) =
2
3(x − 2)(x − 1)
(x − 5)(x + 4)
We’ll have more to say about what continuity really is soon.
2
for all values of x other than 3, and this expression doesn’t blow up at x = 3. So what
happens near x = −4 and x = 5? In both cases, as we choose values of x closer and closer
to the value for which f is undefined, our numerator settles in on some nonzero value, while
our denominator gets closer and closer to zero. But dividing by numbers which are very
close to zero yields numbers of very large absolute value, and thus the graph tends to ±∞.
We call the lines x = −4 and x = 5 vertical asymptotes of f .
Our graph has another asymptote that interests us: as x− > ±∞, we can see that f (x)
is settling in on the value 3. Why? We can try to figure this out by considering the value of
f for larger and larger values of f , but dropping ∞ into our above expression for f is just
going to yield ∞/∞. One common strategy for handling this is to express the numerator
and denominator of f as polynomials, and then divide each term by the highest power of f
which appears in the denominator. That is,
6x3 − 36x2 + 66x − 36
=
f (x) = 3
2x − 8x2 − 34x + 120
6x3
x3
2x3
x3
36x2
+ 66x
− x363
x3
x3
8x2
− 34x
+ 120
x3
x3
x3
−
−
6−
=
2−
36
+ x662 − x363
x
.
8
− x342 + 120
x
x3
Notice that as x becomes very large, most of the terms in this last expression will disappear.
Dividing by larger and larger values will yield values closer and closer to zero, and the only
terms to survive will be the 6 in the numerator and the 2 in the denominator. That is,
lim f (x) =
x→∞
6
= 3.
2
The same analysis will work for x → −∞. Rational functions sometimes have what we call
oblique asymptotes, where their ends approach some line which is not horizontal, but we
won’t explore these here.
Example.
4. Find the vertical and horizontal asymptotes of f (x) =
7x3 − 21x2 + 21x − 7
.
(x − 1)2 (x2 − 7x + 12)
Trigonometric Functions
Given a circle of radius 1 centered at (0, 0), trace out an arclength of θ along the circle. We
call the x-coordinate of the endpoint of this path the cosine of θ, written cos(θ) and call
the y-coordinate the sine of θ, written sin(θ). In this situation, θ is known as the angle
(measured in radians).
The ratio of sin(θ) and cos(θ) is called the tangent of θ:
tan(θ) :=
3
sin(θ)
.
cos(θ)
Figure 1: Some values of the sine and cosine functions.
We also define the reciprocals of these three functions as their own functions:
sec(θ) :=
1
,
cos(θ)
csc(θ) :=
1
,
sin(θ)
cot(θ) :=
1
,
tan(θ)
called secant, cosecant, and cotangent, respectively. (There are far better ways to define
these functions, but it is assumed that you have seen them before.)
Examples.
5. What are the domains and ranges of the trigonometric functions defined above?
6. Find sin(3π/4), cos(10π/3), tan(−π/6), cot(13π/6).
Notice that since cos(θ) is the x-coordinate of our endpoint and sin(θ) is the y-value, we
can obtain a right triangle by moving from the origin to (cos(θ), 0), and then to (cos(θ), sin(θ)).
We can then apply the Pythagorean identity to our triangle to obtain the following important
trigonometric identity:
cos2 (θ) + sin2 (θ) = 1.
(2)
This is a very important identity, from which many other trigonometric identities can be
deduced. A variety of other trig identities can be found in the front cover of your textbook.
Example.
7. Suppose sin(θ) =
and tan(θ).
1
2
and tan(θ) < 0. Use equation (2) to deduce the values of cos(θ)
4
Miscellany
Definition. We call a function f (x) even if f (−x) = f (x) for all x and we call f odd if
f (−x) = −x.
Note. We get the names even and odd for these functions from polynomials. Notice that
f (x) = x2 is even, while g(x) = x3 is odd. Also notice that, unlike the integers, there exist
functions which are neither odd nor even (such as h(x) = x5 + x2 ). What can you say about
a function which is both odd and even?
Example.
8. Are sin(θ) and cos(θ) even, odd, or neither? What about |x|, x1 ,
x2
?
x+1
Remember that we sometimes define functions by using different rules for different parts
of the real line; we call these piecewise-defined functions.
Example.
9. Draw the graph of y = f (x), where

−3 − x,
x < −1
√
2
f (x) :=
1 − x , −1 ≤ x < 1 .

x − 3,
1≤x
5
Solutions to Examples
1. Find the equation of the line passing through (1, 4) with slope 2. Where does this line
intersect the line x = 5?
(Solution) The slope between any point (x, y) on this line and the point (1, 4) should
be 2, so we should have
y−4
=2
x−1
⇒
y − 4 = 2(x − 1)
for all points on the line where x 6= 2. To see that y is a function of x we sometimes
write this as y = 2x + 2. When x = 5 we have y = 12, so the two lines intersect at
(5, 12).
2. Find the equation of the line passing through (−1, 1) and (3, 9). Where does this line
intersect the line y = 13?
(Solution) First, we find the slope of this line to be
8
9−1
= = 2,
3 − (−1)
4
so the line is determined by the equation
y − 9 = 2(x − 3),
or y = 2x + 3. When y = 13 we have 13 = 2x + 3, so x = 5. The lines intersect at
(5, 13).
3. For what values is the rational function f (x) =
(6x − 12)(x2 − 4x + 3)
defined?
(2x − 6)(x2 − x − 20)
(Solution) This function is defined whenever the denominator is not zero, so we find
the zeros of the denominator. One way to do this is by factoring:
(2x − 6)(x2 − x − 20) = 2(x − 3)(x − 5)(x + 4).
Having factored, we see that the denominator vanishes when x = −4, x = 3, and
x = 5. So f (x) is defined for all x-values other than these three.
4. Find the vertical and horizontal asymptotes of f (x) =
6
7x3 − 21x2 + 21x − 7
.
(x − 1)2 (x2 − 7x + 12)
(Solution) First we’ll factor the polynomials involved:
f (x) =
7(x − 1)3
7x3 − 21x2 + 21x − 7
=
.
(x − 1)2 (x2 − 7x + 12)
(x − 1)2 (x − 3)(x − 4)
For values other than x = 1, this reduces to
f (x) =
7(x − 1)
.
(x − 3)(x − 4)
(3)
Now f (x) is undefined for x = 1, x = 3, and x = 4, so these are our “candidates” for
vertical asymptotes. For x-values very close to (but not exactly) 1, we can use (3) to
compute f (x), and we see that f (x) is settling in on 0 as our x-values get closeer to 1.
So f is undefined at x = 1, but doesn’t have an asymptote. For x = 3 and x = 4, on
the other hand, choosing x-values near these values will cause f to blow up in absolute
value, indicating that we have vertical asymptotes at these two points.
For a horizontal asymptote, we investigate the behavior of f as x tends to ∞ in
asbsolute value. We again use our strategy of dividing each term by the highest power
of x that appears in the denominator. We have
7x3 − 21x2 + 21x − 7
x→∞ x4 − 9x3 + 27x2 − 31x + 12
2
7x3
7
− 21x
+ 21x
4 − 4
x4
x4
= lim x4 9x3 27x2 x 31x x 12
x→∞
− x4 + x4 − x4 + x4
x4
lim f (x) = lim
x→∞
= lim
x→∞
7
x
1−
−
9
x
21
x2
+
+
27
x2
21
x3
−
−
31
x3
7
x4
+
12
x4
.
As before, the growing denominators will send most terms to zero, so that
lim f (x) =
x→∞
0−0+0−0
= 0.
1−0+0−0+0
A similar analysis applies for x → −∞, so we see that f has a horizontal asymptote
at y = 0.
5. What are the domains and ranges of the trigonometric functions defined above?
(Solution) For sine and cosine, we can plug in any angle we like; angles greater than
2π just involve winding around the circle, and negative angles entail walking about the
circle “backwards”, or clockwise. So sine and cosine have R as their domain. Notice
that neither the x- nor y-value of a point on the unit circle can be greater than 1 or less
than -1, so both sine and cosine have a range of [−1, 1]. We have to be more careful
with the remaining four functions. For example, tangent is given as sine divided by
7
cosine, so the tangent is not defined whenever cosine vanishes. Since cos θ = 0 whenever
θ = kπ/2 for some odd integer k, we see that the domain of tangent is all real values
not of this form. The tangent can be arbitrarily large or small, however. For instance,
near θ = π/2, sin θ is close to 1 and cos θ is close to 0, so tan θ is a very large, positive
number. Near θ = −π/2 on the other hand, sin θ is close to -1 and cos θ is close to 0,
so tan θ is a very negative number. The range of tangent is all real numbers. Similarly,
one may find that secant has the same domain as tangent, and that the range of secant
is (−∞, −1] ∪ [1, ∞). The domain of cosecant is all real numbers not of the form nπ
for some integer n, and the range of cosecant is the same as the range of secant. The
domain of cotangent is the same as the domain of cosecant, and the range of cotangent
is all nonzero numbers.
6. Find sin(3π/4), cos(10π/3), tan(−π/6), cot(13π/6).
√
(Solution) We can read sin(3π/4) directly off the unit circle: sin(3π/4) = 2/2. For
cos(10π/3), we√use the fact that cos(θ + 2π) = cos(θ) for all angles θ, so cos(10π/3) =
tan(−π/6) = tan(11π/6). At 11π/6 we have sin(11π/6) =
cos(4π/3) = − 2/2. Similarly,
√
−1/2 and cos(11π/6) = 3/2, so
−1/2
1
tan(−π/6) = √
= −√ .
3/2
3
Finally, cot(13π/6) = cot(π/6), so
√
cos(π/6)
1
3/2 √
=
=
= 3.
cot(13π/6) =
tan(13π/6)
sin(π/6)
1/2
7. Suppose sin(θ) =
and tan(θ).
1
2
and tan(θ) < 0. Use equation (2) to deduce the values of cos(θ)
(Solution) We have sin2 (θ) + cos2 (θ) = 1, so
cos2 (θ) = 1 − sin2 (θ) = 1 −
1
3
= .
4
4
√
This means that cos(θ) =√± 3/2. But since tan(θ) < 0 and sin(θ) > 0, we must have
cos(θ) < 0, so cos(θ) = − 3/2. Then
tan(θ) =
1/2
1
√
= −√ .
− 3/2
3
We can look at our unit circle and see that if θ ∈ [0, 2π), then θ = 5π/6.
8. Are sin(θ) and cos(θ) even, odd, or neither? What about |x|, x1 ,
8
x2
?
x+1
(Solution) Consider replacing θ with −θ in sin(θ). This amounts to rotating clockwise
from (1, 0) instead of counter-clockwise, and we see that the vertical position of the
resulting point on the unit circle is the negative of the vertical position following a
counter-clockwise rotation. That is,
sin(−θ) = − sin(θ),
so sine is an odd function. Note that rotating clockwise does not affect the horizontal
position, so cos(−θ) = cos(θ), meaning that cosine is an even function.
For the remaining functions, we replace x with −x and check the resulting behavior.
Since
| − x| = |x|,
we see that the absolute value function is even. On the other hand,
1
1
=−
,
−x
x
so f (x) = 1/x is an odd function. Finally,
x2
x2
(−x)2
=
6= ±
,
(−x) + 1
1−x
x+1
so the function g(x) = x2 /(x + 1) is neither odd nor even.
9. Draw the graph of y = f (x), where

−3 − x,
x < −1
√
2
f (x) :=
1 − x , −1 ≤ x < 1 .

x − 3,
1≤x
(Solution) This graph is obtained by graphing each of the three listed functions on
their respective domains. Here’s a plot produced by Mathematica:
9