Transcript No Slide Title

Vertical and horizontal shifts
• If f is the function y = f(x) = x2, then we can plot points and
draw its graph as:
y
x
• If we add 1 (outside change) to f(x), we have y = f(x) + 1 =
x2 + 1. We simply take the graph above and move it up 1 unit
to get the new graph.
y
 (2,5)
x
• If we replace x by x+2 (inside change) to form the function
y = f(x+2) = (x+2)2, then the corresponding graph is obtained
from the graph of y = x2 by moving it 2 units to the left along
y
the x-axis.
x
(-2,0)
• If we replace x by x–1 (inside change) to form the function
y = f(x–1) = (x–1)2, then the corresponding graph is obtained
from the graph of y = x2 by moving it 1 unit to the right along
the x-axis.
y
(1,0)
x
If y = g(x) is a function and k is a constant, then the graph of:
• y = g(x) + k is the graph of y = g(x) shifted vertically by |k|
units. If k > 0, the shift is up, and if k < 0, the shift is down.
• y = g(x+k) is the graph of y = g(x) shifted horizontally by |k|
units. If k > 0, the shift is left, and if k < 0, the shift is right.
Horizontal and vertical shifts of the graph of a function are
called translations.
An example which combines horizontal and vertical shifts
• Problem. Use the graph of y = f(x) = x2 to sketch the
graph of g(x) = f(x–2) – 1 = (x–2)2 – 1.
Solution. The graph of g is the graph of f shifted to the
right by 2 units and down 1 unit as shown below.
y
(2,-1)
x
Reflections and symmetry
• Suppose that we are given the function y = f(x) as shown.
y
x
• If we define y = g(x) = –f(x), then the graph of g may be
obtained by reflecting the graph of f vertically across the
x-axis as shown next.
y
x
Continuation of example from previous slide
• If we define y = h(x) = f(–x), then the graph of h is obtained
by reflecting the graph of f horizontally across the y-axis as
shown next.
y
x
• Next, we define y = p(x) = –f(–x). The graph of p is obtained
by reflecting the graph of f about the origin as shown next.
y
x
For any function f:
• The graph of y = –f(x) is the reflection of the graph of
y = f(x) across the x-axis.
• The graph of y = f(–x) is the reflection of the graph of
y = f(x) across the y-axis.
• The graph of y = –f(–x) is the reflection of the graph of
y = f(x) about the origin. Note that this reflection can be
obtained by applying the two previous reflections in
sequence.
Symmetries of graphs
• A function is called an even function if, for all values of x in
the domain of f,
f( x)  f(x).
The graph of an even function is symmetric across the y-axis.
Examples of even functions are power functions with even
exponents, such as y = x2, y = x4, y = x6, ...
• A function is called an odd function if, for all values of x in
the domain of f,
f( x)   f(x).
The graph of an odd function is symmetric about the origin.
Examples of odd functions are power functions with odd
exponents, such as y = x1, y = x3, y = x5, ...
• Problem. Is the function f(x) = x3+x even, odd, or neither?
Solution. Since –2 = f(–1) is not equal to f(1) = 2, it follows that
f is not even.
Since f(–x) = (x)3  (x)  x 3  x = –f(x), it follows that f is
odd.
y = x3+x
Note the symmetry
about the origin.
• Problem. Is the function f(x) = |x| even, odd, or neither?
Solution. Since f(–x) = |x| = f(x), it follows that f is even.
Since 1 = f(–1) is not equal to –f(1) = –1, it follows that f is not
odd.
y = |x|
Note the symmetry
about the y-axis.
• Question. Is it possible for a function to be both even and odd?
Combining shifts and reflections--an example
• In an earlier example, we discussed an investment of
$10000 in the latest dotcom venture. This investment had
a value of 10000(0.95)t dollars after t years. Suppose that
we want to graph the amount of the loss after t years for
this investment. The formula for the loss is:
10000 – 10000(0.95)t
Shift Upwards Reflect across t-axis
• The loss is graphed on the next slide using Maple.
Use of Maple to graph loss on dotcom investment
> plot({10000,10000-10000*(0.95)^t},t=0..80, color=black,labels=["t","L"]);
The graph of the loss has a horizontal asymptote, L = 10000.
Vertical Stretches and Compressions
• If f(x) = x2 and g(x) = 5x2, then the graph of g is obtained
from the graph of f by stretching it vertically by a factor of
5 as the following Maple plot shows:
• If f(x) = x2 and g(x) = -5x2, then the graph of g is obtained
from the graph of f by stretching it vertically by a factor of
5 and then reflecting it across the x-axis as the following
Maple plot shows:
• If we compare the graphs of f(x) = x2 and g(x) = (1/2)x2,
we notice that the graph of g can be found by vertically
compressing the graph of f by a factor of 1/2.
• Generalizing the above examples yields the following:
If f is a function and k is a constant, then the graph of
y = kf(x) is the graph of y = f(x)
• Vertically stretched by a factor of k, if k > 1.
• Vertically compressed by a factor of k, if 0 < k < 1.
• Vertically stretched or compressed by a factor |k|
and reflected across the x-axis, if k < 0.
Vertical Stretch Factors and Average Rates of Change
• If f(x) = x2 and g(x) = 5x2, we compute the average rates of
change of the two functions on the interval [1,3] as follows:
 Average rate of change  Change in f(x) f(3) – f(1)

 4.
over the interval 1  x  3 
Change in x
3 –1


 Average rate of change  Change in g(x) g(3) – g(1)

 20.
over the interval 1  x  3 
Change in x
3 –1


• The above computation illustrates a general fact:
If k is constant and g(x)  k  f(x), then on any interval,
Average rate of change of g  k  (Average rate of change of f).
• If f(x) = 4–x2 and g(x) = 4 – (2x)2, then the graph of g is
obtained from the graph of f by compressing it horizontally
by a factor of 1/2 as the following Maple plot shows:
• If f(x) = 4–x2 and g(x) = 4 – (0.5x)2, then the graph of g is
obtained from the graph of f by stretching it horizontally
by a factor of 2 as the following Maple plot shows:
• Generalizing the two previous examples yields the
following results for horizontal stretch or compression.
• If f is a function and k is a positive constant, then the graph
of y = f(kx) is the graph of f
• Horizontally compressed by a factor of 1/k if k > 1.
• Horizontally stretched by a factor of 1/k if k < 1.
If k < 0, then the graph of y = f(kx) also involves a
horizontal reflection about the y-axis.
The Family of Quadratic Functions
• A quadratic function is a function with a formula in one
of the following forms:
• Standard form: y = ax2+bx+c, where a, b, c, are
constants, a  0.
• Vertex form: y = a(x–h)2+k, where a, h, k are
constants, a  0.
• The graph of a quadratic function is called a parabola.
• Conversion from one form to the other for a quadratic
function is discussed on the next slide.
• To convert a quadratic function from vertex form to
standard form, simply multiply out the squared term. To
convert from standard form to vertex form, we complete
the square as illustrated in the following example.
• Example. Put the following quadratic function into vertex
form by completing the square.
t(x)   4x 2  12x  8.
(1) Factor out the coefficient of x2, which is – 4.
t(x)   4(x 2  3x  2).
(2) Add and subtract the square of half the coefficient of
the x-term.
9 9


t(x)   4 x 2  3x    2 .
4 4


perfect square
(3) Write the equation in vertex form.
2
3

t(x)   4 x    1.
2

The Vertex of a Parabola
• Recall that the graph of a quadratic is called a parabola.
• The parabola corresponding to y = a(x–h)2+k:
• Has vertex (h, k).
• Has axis of symmetry x = h.
• Opens upward if a > 0 or downward if a < 0.
Finding the vertex of a parabola
• Example. For the previous example, graph the parabola
and find the vertex.
2
3

t(x)   4 x    1.
2

We note that the vertex is at (–3/2, 1). The graph
follows:
Finding a formula for a parabola
• If we know the vertex of a quadratic function and one
other point, we can use the vertex form to find its formula,
as shown in the following example.
• Example. A parabola has vertex at (–3, 2) and (0, 5) is on
the parabola. Find the formula for the corresponding
quadratic, f(x). Use the vertex form with h = –3 and k =2.
This results in
f(x)  a(x  (3)) 2  2, or
f(x)  a(x  3) 2  2.
To find the value of a, we substitute x = 0 and y = 5 into
this formula, obtaining a = 1/3. The formula is therefore
1
2
f(x)  x  3  2.
3
Finding a formula for a parabola, continued.
• If three points on a parabola are given, we can use the standard
form of the corresponding quadratic to find the formula.
• Example. Suppose the points (0, 6), (1, 0), and (3, 0) are on a
parabola. Find a formula for the parabola. Use the standard
form: y = ax2+bx+c. Since (0, 6) is on the parabola, it follows
that c = 6. From the other two points, we have:
0  a(1) 2  b(1)  6
0  a(3) 2  b(3)  6.
This system can be solved simultaneously for a and b. We
obtain a = 2 and b = –8. Thus, the equation of the parabola is
y = 2x2–8x+6.
Finding the zeros of a quadratic function.
• The zeros of a function f are values of x for which f(x) = 0.
• In addition to the standard and vertex forms, some quadratic
functions f(x) can also be expressed in factored form:
f(x)  a(x  r)(x  s), where a, r, and s are constants, a  0.
• Example. Find the zeros of f(x) = x2–x –6. Set f(x) = 0 and
solve for x. We have x2–x –6 = 0. We next express f(x) in
factored form, so it will be easy to find the zeros.
(x  3)(x  2)  0.
The zeros are x = 3 and x = –2. Note that these are the values r
and s from the factored form.
Finding a formula for a parabola using the factored form
• Example. Suppose the points (0, 6), (1, 0), and (3, 0) are on
a parabola, as in a previous example. Find a formula for
the parabola using the factored form. Since the parabola
has x-intercepts at x = 1 and x = 3, its formula is
y  a(x  1)(x  3).
Substituting x = 0, y = 6 gives 6 = 3a or a = 2. Thus, the
equation is:
y  2(x  1)(x  3).
If we multiply this out, we get y = 2x2–8x+6, which is the
same result as before.
Two methods for finding the zeros of a quadratic
• The first method involves completing the square. Suppose
we want the roots of x2 + 3x + 2 = 0. If we complete the
square as before, we get (x + 1.5)2– 0.25 = 0. If we rewrite
this as (x + 1.5)2 = 0.25, we can take the square root of
both sides of the equation to get x + 1.5 = 0.5, which
gives x = –1 and x = –2.
• The other method involves the use of the quadratic
formula, which was presented in a previous slide lecture.
If we apply the quadratic formula to x2 + 3x + 2 = 0, we get
x
– 3  3 2  4(2)
2
This reduces to x = –1.5  0.5 . Again, x = –1 and x = –2.
• What does it mean if a quadratic does not have real zeros?
It means that the graph of the corresponding parabola does
not cross the x-axis.
• Problem. If we have 4 feet of string, what is the rectangle
of largest area which we can enclose with the string?
Solution. If we let one side of the rectangle have length x,
then the other side must have length (4–2x)/2. That is, the
other side is 2–x. Therefore, the area of the rectangle is
a(x) = x(2–x) = –x2+2x. If we write this in vertex form, we
have a(x) = –(x–1)2+1. Thus, the vertex is at (1, 1), and
the rectangle of maximum area is a square with a side
length of 1 foot.
Summary for Transformation of Functions and their Graphs
• If y = g(x) is a function and k is a constant, then the graph of
y = g(x) + k is the graph of y = g(x) shifted vertically by |k| units.
• If y = g(x) is a function and k is a constant, then the graph of
y = g(x+k) is the graph of y = g(x) shifted horizontally by |k| units.
• A function is called an even function if, for all values of x in the
domain of f, f(–x) = f(x). The graph of an even function is
symmetric across the y-axis.
• A function is called an odd function if, for all values of x in the
domain of f, f(–x) = –f(x). The graph of an odd function is
symmetric about the origin.
Summary for Transformation of Fcts and their Graphs, cont’d
• When a function f(x) is replaced by kf(x), the graph is
vertically stretched or compressed and the average rate of
change on any interval is also multiplied by k. If k is negative,
a vertical reflection about the x-axis is also involved.
• When a function f(x) is replaced by f(kx), the graph is
horizontally stretched or compressed by a factor of 1/|k| and,
if k < 0, reflected horizontally about the y-axis.
• A quadratic function has a formula in either standard form or
vertex form. Completing the square converts standard to vertex
form. Vertex form is used to find the max or min value of the
quadratic function. A quadratic function has 0, 1, or 2 real
zeros. If a quadratic function has real zeros, it can also be
represented in factored form. Methods for finding the formula
for a quadratic function from given data points were discussed.