Transcript Quad Functions and Models

f x   x 2
f x  
f x   3x 2
f x   ax
1 2
x
3
If a < 0 the parabola opens down
and the larger the a the
“narrower” the graph and the
smaller the a the “wider” the
graph.
2
If a > 0 the parabola opens up
and the larger the a value the
“narrower” the graph and the
smaller the a value the “wider”
the graph.
1
f x    x 2
3
f x    x 2 f x   3x 2
The graph of
this function is
a parabola
f x   ax  h   k
2
Determines whether
the parabola opens
up or down and how
“wide” it is
horizontal shift,
moves graph
horizontally by h
vertical shift,
moves graph
vertically by k
Thisneed
We
will factor
to algebraically
into (x-3)(x-3) so
manipulate
we
can express
this to
it as
look
something
like the
Let’s look squared
at a above.
form
and We’ll
combine
do this
the by
-1 and
quadratic -9
completing
on the end.
the square.
function and see
2
if we can graph
it.
9
9
f x   x  6 x  1
2
f x  
 101  ____
f xx  
 6xx 3___
2
Subtract it here to keep things equal
Add a number here to
make a perfect square
(can’t add a number without compensating
for it and we don’t want to add it to the
other side because of function notation)
down 10
f x   x  3  10
2
right 3

We started with f x  x 2  6 x  1 and completed the
square to get it in the format to be able to graph using
transformations.
We can take the general quadratic equation and do this to
find a formula for the vertex.
What we find from doing this is on the next slide.
f x   ax  bx  c
2
Let’s try this on the one we did before:
2
f x  1x  6 x  1
The x value of the vertex of the parabola can be
found by computing  b
2a
 (-6)
b
x value of vertex 
2a(1)
3
The vertex is
then at (3, -10)
The y value of the vertex of the parabola can be
found by substituting the x value of the vertex in
the function and finding the function value.
y value of vertex  f 3  3  63  1  10
2
Let’s plot the vertex:
(0, -1)
(6, -1)
Since the a value is positive, we
know the parabola opens up.
The parabola will be
symmetric about a vertical
line through the vertex called
the axis or line of symmetry.
ffxx0x 660x11
22
(3, -10)
Let’s find the y intercept by
plugging 0 in for x.
So y intercept is (0, -1)
We can now see enough to
graph the parabola
The graph is symmetric with
respect to the line x = 3 so
we can find a reflective
point on the other side of
the axis of symmetry.
Let’s look at another way to graph the parabola starting
with the vertex:
We could find the x intercepts
of the graph by putting f(x)
(which is the y value) = 0
f0x x x 6x6x1 1
2 2
(3, -10)
This won’t factor so we’ll
have to use the quadratic
formula.
  6   6  41 1
 b  b  4ac

x
21
2a
6  40

 6.2 and  0.2 So x intercepts are (6.2, 0)
and (- 0.2, 0)
2
2
2
A mathematical model may lead to a quadratic function.
Often, we are interested in where the function is at its
minimum or its maximum. If the function is quadratic the
graph will be a parabola so the minimum (if it opens up) will
be at the vertex or the maximum (if it opens down) will be at
the vertex.
b
We can find the x value of the vertex by computing 
2a
We could then sub this value into the function to find its
minimum or maximum value.
DEMAND EQUATION
The price p and the quantity x sold of a certain product obey
the demand equation:
1
p   x  200, 0  x  400
2
Express the revenue R as a
function of x.
Revenue is the amount you bring in,
so it would be how much you charge
(the price p) times how many you sold
(the quantity x)
This is the real world domain.
The equation doesn’t make
sense if the quantity sold is
negative (x < 0) and it doesn't
make sense if the price is
negative (if x > 400)
R  xp
1 2
 1

R  x  x  200    x  200 x
2
 2

1 2
R   x  200 x
2
This is a quadratic equation and
since a is negative, its graph is a
parabola that opens down. It will
have a maximum value then at the y
value of the vertex.
What is the revenue if 100 units are sold?
1
2
R   100  200100  $15,000
2
b
200

 200
What quantity x maximizes revenue? x  
2a
 1
2  
 2
Since the revenue function is maximum at the vertex, we'll want to
find the x value of the vertex to answer this.
1
2
What is the maximum revenue? f 200   200  200200
2
This would be the y value of the vertex
 $20,000
DEMAND EQUATION
The price p and the quantity x sold of a certain product obey
the demand equation:
1
p   x  200, 0  x  400
2
1 2
R   x  200 x
2
What price should the company charge to receive
maximum revenue?
Since we just found that the quantity to achieve maximum
revenue was 200, we can substitute this in the price equation
to answer this question.
1
p   200   200  $100
2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au