Polynomial, Power, and Rational Functions by bfs11840

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									Polynomial, Power, and
  Rational Functions
        Chapter 2
 Linear and Quadratic
Functions and Modeling
          2.1
Polynomial
             Polynomials
A polynomial, in other words, cannot have x
  raised to a fractional, zero, or negative
  power. It also cannot have x in the power.
 Is the Function a Polynomial?
F(x)=-9 + 2x
Yes since no powers of x are fractional and
  x is not a power.
F(x)=13
Yes for the same reason as above
F(x)=3x
No, because x is not the base it is the
  power.
  Is the Function a Polynomial?
F(x)= x1/2
No, because x is raised to a fractional
  power.
          Linear Function
A linear function can be written in the form
                  Y=mx+b
     Finding a Linear Function
1. Find the coordinates.
2. Find the slope or the average rate of change by
   finding the change in y over change in x.
3. Plug the slope in for m in either y=mx+b or y-y1
   = m(x-x1).
4. Plug in one of the points for x and y or x1 and
   y1, and then solve for b or y.
5. Write out final equation with m and b plugged
   in.
    Linear Function, Example
Find the equation of a linear function given:
F(-3) = 5 and f(6) = -2
   Graphing a Linear Function
If given two points, then plot each point and
   connect.
Standard Form
              i



Remember that x and y
intercepts can help you
decide which graph is the
graph of the given quadratic
function.
Match a Parabola to its Function
        in Vertex Form
1. Find the vertex and look for the graph that
   has that vertex.
2. Determine if a is positive or negative, so
   the graph should point up or down.
3. Find the x and y intercepts and check
   them with the graph.
 Transformations of a Parabola
Remember:
-aF(-b(x-h)) + k
- Flips over the x-axis
A—vertical stretch if a>1, shrink if 0<a<1
- Flips over the y-axis
B—horizontal stretch if 0<b<1,shrink if b>1
H—moves left/right
K—moves up/down
 Transformations of a Parabola
¼ x^2 – 1
The ¼ is a and so there is a vertical shrink by ¼.
The ¼ is positive, so the graph does not flip over
  the x –axis.
There is no number being added to or subtracted
  from x, so the graph does not move left or right.
The x is not negative so the graph does not flip
  over the y-axis.
The -1 moves the graph down 1.
    Vertex of a Parabola

The point of the
 parabola that is the
 max or the min.
   Find the Vertex and Axis of
           Symmetry
G(x) = -3(x+2)^2 – 1
Vertex is (h,k) or (-2,-1)
Axis of symmetry is x=h or x= -2
How to put a Quadratic Function
        in Vertex Form
Use Completing the Square
  Using Completing the Square to
  Rewrite a Quadratic Function in
           Vertex Form
Put y or f(x) on the left side of the equation
 and all other terms on the other side of the
 equation. Simplify the right side if needed.

I will use this example to show what the
  steps mean. We will rewrite Y=3x2+5x-4
  in vertex form, which is Y = a(x-h)2 +k.
  Using Completing the Square to
  Rewrite a Quadratic Function in
           Vertex Form
Factor out the coefficient of the x^2 term, a,
  unless it is one, from all the terms with x
  on the right side of the equation.
In our example, since a is 3 we need to
  factor a three out from the x^2 and the x
  terms.
Y=3(x2+5/3x)-4
Y = a(x-h)2 +k
  Using Completing the Square to
  Rewrite a Quadratic Function in
           Vertex Form
Now, divide the coefficient of the x term by
  two and then square the result.
(5/3)/2 = 5/6     (5/6)2=25/36
  Using Completing the Square to
  Rewrite a Quadratic Function in
           Vertex Form
Add the resulting number to the term in the
  parenthesis. Note that you are adding ―a‖ times
  this amount to the function, so you will then need
  to subtract this same amount so that the function
  is not changed.
In our example we are adding 25/36 to the function
  3 times, so we will then subtract 3*25/36.
Y=3(x2+5/3x+25/36)-4-3*25/36
  Using Completing the Square to
  Rewrite a Quadratic Function in
           Vertex Form
Factor the quadratic part and simplify the
  constant part and the function will be in
  vertex form.
Y=3(x + 5/6)2-73/12
Y = a(x-h)2 +k
Write an Equation for the Parabola
  Given the Vertex and a Point
Use y=a(x-h)^2 + k to help.
1. Plug the x value of the vertex in as h and
   the y value in as k.
2. Take the other coordinate and plug in x
   for x and y for y.
3. Solve this equation for a.
4. Write out the vertex form for the parabola
   with the a, h, and k plugged in.
                 Modeling
• Enter and plot the data as a scatter plot.
• Find the regression model that best fits the
  problem situation. Linear vs. quadratic
  – Does the scatter look linear or curved?
• Place the regression equation into y1, and
  observe the fit to double check.
• Use the regression model to make the
  predictions called for in the problem.
Describing a Linear Correlation
• If the scatter points look like they are
  clustered along a line, then they have a
  linear correlation.
  – Positive linear correlation—positive slope
  – Negative linear correlation—negative slope
• r – correlation coefficient—measures the
  strength and direction of the linear
  correlation of the data set.
    Linear Correlation Coefficient
             Properties
• [-1,1]
• r>0 there is a positive linear correlation
• r<0 there is a positive linear correlation
• |r|=1, then there is a strong linear
  correlation.
• r=0, there is a weak or no linear correlation
  Describing the Strength and
Direction of a Linear Correlation
1. State strong or weak based on how linear
   the points look. The more linear they
   look, the stronger the linear correlation.
2. Then state positive or negative for
   direction. Positive slope or negative
   slope.
       Quadratic Modeling

• Finding maximums and
  minimums
 –Calculator
   • Use max/min function
 –By hand
   • Find the vertex and determine if it is a
     max or a min by determining which
     way the graph points.
          Quadratic Modeling
Creating Vertical Free-Fall Motion
  equations.
S(t) = -1/2 gt2 +vot+so
v(t) = - gt +vo
t is time, g=32ft/s2 or 9.8m/s2 is the acceleration
   due to gravity.
Initial height—so
Initial velocity--vo
  Quadratic Modeling Example
Fireworks are shot by remote control into the
  air from a pit that is 10 ft below the earth’s
  surface.
Find an equation that models the height of
  an aerial bomb t seconds after it is shot
  upwards with an initial velocity of 80ft/sec.
        Example Continued
We need to find a distance equation.
S(t) = -1/2 gt2 +vot+so
so = 10
vo = 80
g = 32 ft/sec^2
S(t) = -1/2 (32)t2 +80t+10
S(t) = -16t2 +80t+10
        Example Continued
What is the maximum height above ground
 level that the aerial bomb will reach? How
 many seconds will it take to reach that
 height?
The maximum height will be at the peak of
 the parabola or the vertex. So use the
 max function on the calculator and the y
 value will be the max height and the x will
 be the time it takes to reach the height.
        Example Continued
What is the maximum height above ground
 level that the aerial bomb will reach? How
 many seconds will it take to reach that
 height?
By hand, find –b/2a and this will be the time
 it takes to reach the height. Plug this
 value into the function and the result will
 be the max height.

								
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