Math 110
Chapter 3, Section 1
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Review Notes From ’95:
f ( x)  ax  h   k
Since k gets added to the thing directly, it shifts the parabola up if it's positive, down if negative
The axis of symmetry is h
negative h shifts the parabola right
positive h shifts the parabola left
The larger a is, the narrower the parabola
the smaller a is, the broader
If a is positive, it opens upwards
If a is negative, it opens downwards
2
Vertex at (h,k)
So if this is so convenient, how do change stuff into this format?
Perfect squares!
f ( x)  x 2  6 x  10
2
b
Add (and subtract)   to the right side, b is -6
 2
2
f ( x)  x  6 x  9  9  10
f ( x)  x  3  1
2
Vertex is at (3,1), opens upwards, get the y-intercept by plugging x
=0
Math 110
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Math 110
Chapter 3, Section 1
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Review: Quadratic Functions
Axis of symmetry: for f ( x)  ax 2  bx  c the axis of symmetry is about x, where x  
b
2a
Explain the symmetry idea
The vertex is either the highest or lowest point of the parabola
You get it by plugging in the x value of the axis of symmetry, and
4ac  b 2
you get out
4a
Ex: Given y   x 2  8x  12 y
a) Does the parabola open upward or downward?
b) Find the y-intercept
c) Find the vertex
d) Find the x-intercepts, if any
e) Draw the Graph
a) -1  downward
b) set x = 0, and solve. (The y-intercept is where the function crosses the Y axis – where x
is zero. Therefore, evaluating f(0) gets us the answer)
 b 4ac  b 2 

c)   ,
2
a
4
a


d) set y = 0, and solve
Notice that if the parabola is entirely above the x axis, there will be no real solutions.
If it touches the x axis in exactly one place, there will be one.
If it crosses the x axis, it’ll have 2.
We can use the discriminant to figure this out ahead of time:
 b  b 2  4ac
discriminant = b  4  a  c (assuming that f x  
)
2a
If discriminant > 0, then 2 real solutions
If discriminant = 0, then 1 real solutions
If discriminant < 0, then no real solutions
2
e) Draw the graph based on the above
<Do some problems based around this>
Math 110
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Math 110
Chapter 3, Section 1
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Using Quadratic Functions To Model Problems
We can now solve problems of the sort 'find the maximum value' or 'find the min' value
Given a quadratic function
Find the axis of symmetry – if it opens
up, that's the min, otherwise it's the
max
N  p  490  7 p
R  x p
R p  N  p  p
R p  490  7 p  p
R p   490 p  7 p 2
Next, find the price that maximizes the revenue, by using the fact that the max is at the vertex
b
b
4ac  b 2
x
tells me where the max is, y 
tells me what the max is (alt, plug 
directly
2a
2a
4a
into R(p) & get the result out that way
Having gotten the maximized price, we can then plug that back into N(p) to find how many units
we'll need to sell, and into R(p) to find out what the max revenue is.
Quadratic Regressions
Last lecture we saw the basic overview of how to model real-world stuff (Get the real-world
data, draw scatter diagrams, do the regression)
 Not every set of data you get will be linear (as a matter of fact, most won’t)
 So part of your job in the “do the regression” step will be to (intelligently) try out different
types of functions (linear, quadratic, etc), and see which one fits best.
Run through the quad. regression procedure on the TI-83
 MAKE SURE TO POINT OUT THAT YOU NEED TO TURN DIAGNOSTICS ON!!!
o Also point out that the only way to do this is through the “DiagnosticsOn” command,
picked directly off the catalog (on the TI-83, at least)
<Have them first identify which scatter plots can be modeled by functions that are linear, are
quadratic, and are something else entirely>
Math 110
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