quadratic functions (3.1). identifying the vertex (e2, p243) complete the square
TRANSCRIPT
Quadratic Functions (3.1)
12102function theofMin/Max
2at vertex 2
122
27222
272242
2742
782
2
22
222
2
2
x
x
x
xx
xx
xx
Identifying the vertex (e2, p243) Complete the square
110 function ofmin/max ;3at vertex
13
833
8336
86
86
2
22
222
2
2
x
x
x
xx
xx
xx
Identify both the vertex and the x intercept (e3, p244)
intercepts axisx 4,2)4)(2(862 xxxxxFactor it
Given vertex and a point find the equation of a parabola (e4) (p244) plug in all 4 values given.
212
22136
),(
2
2
2
xy
aa
khvertexkhxay
Higher Degree Polynomials (3.2)
negative coefficient reflects the graph in the x-axi
Degree is odd
upward shift, by one unit left shift, by one unit
the degree is odd and the leading coefficient is negative, the graph rises to the left and falls to the right
the degree is even and the leading coefficient is positive, the graph rises to the left and right
the degree is odd and the leading coefficient is positive, the graph falls to the left and rises to the right
The Leading Coefficient Test only tells you whether the graph eventually rises or falls to the right or left. Other characteristics of the graph, such as intercepts and minimum and maximum points, must be determined by other tests.
Page 255
Page 255
Apply Leading Coefficient Test. Because the leading coefficient is positive and the degree is even, you know that the graph eventually rises to the left and to the right
2346 23 xxxxfExample 11 (P274): Find the zeros of
Step1: plot the graph – let the calculator/computer do the work
There is a zero here, between 0.6 and .07
Step2: Rational Zero Test (P270)
6
1,
3
2,
3
1,
2
1,2,1
6,3,2,1
2,1
6 of factors
2 of factors
Setp3: Test
023
233243
26
234623
23
xxxxf Plug it into the calculator, don’t try to evaluate it by hand
Yes 2/3 is a zero
x3 x2 x c
2/3 6 -4 3 -2
4 0 2
6 0 3 0
Setp4: Synthetic Division
363
2 2
xxxf
Remainder is 0 Another proof that 2/3 is a zero
6(2/3)
Remainder and factor theorems on page 268; e5,6
We will skip the upper and lower bound rule on page 258
6x2 3
x-2/3 6x3 -4x2 3x -2
6x3 -4x2
3x -2
0
Now the long division is much easier (p264, e1,2,3)
Example 3 (p28): 8122 235 xxxxxf
Possible zeros, repeated (touches)?
A zero
8,4,2,11
8,4,2,1
1 of factors
8 of factors
Step1: Plot it
Step2: Rational Zero Test (p 256)
x5 x4 x3 x2 x c
-2 1 0 1 2 -12 8
-2 4 -10 16 -8
1 -2 5 -8 4 0
48522 234 xxxxx
x4 x3 x2 x c
1 1 -2 5 -8 4
1 -1 4 -4
1 -1 4 -4 0
4412 23 xxxxx
x3 x2 x c
1 1 -1 4 -4
1 0 4
1 0 4 0
4)1(12 2 xxxx
Setp3: Test: Plug in 1 and 2
Setp4: Synthetic Division
)2)(2)(1(12
2)1(124)1(12 222
ixixxxx
ixxxxxxxx
Example 1, p286
100001.
01.100
1100
1)(
f
f
xxf
As x (input) gets bigger y (output) gets smaller
As x (input) gets smaller y (output) gets larger
0
domain
asymptotea straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the curve at the point approaches the slope of the line
Rational Functions and Asymptotes (3.5)
Degree of the numerator is equal to the degree of the denominator Horizontal asymptote: y= ratio of leading coefficients
Degree of the numerator is less than the degree of the denominator Horizontal asymptote: y = 0
Vertical asymptotes: set the denominator equal to zero and solve the resulting equation for x
Degree of the numerator is greater than the degree of the denominator No horizontal asymptote
Example 4 (p289)
Example 5 (p290)
Degree of numerator < denominator Horizontal asymptote y = 0
0 intercept y 00
0 set x to
f
0
02
0 y to
2
xxx
x
set
Example 3 (p298)
x2 x c
-1 1 -1 0
-1 2
1 -2 2
1
22
1
221
1
2
x
xx
xx
x
xxxf
Synthetic Division
Slant Asymptotes – Page 299
Slant asmyptote
1
2
1
21
1
22
xx
x
xx
x
xxxf
x2 x c
1 1 -1 -2
1 0
1 0 -2
Example 5 (p299)
Slant asmyptote