Transcript Slide 1
Quadratic Functions (3.1)
Identifying the vertex (e2, p243) Complete the square 2
x
2 2
x
2 8
x
2
x
2 2
x
x
2 2 2 4 2 4 2
x x
7 2 2 7 2 1 2 2 2 7 vertex at
x
2 2 2 2 7 2 Min/Max of the function 2 0 1 2 1
Identify both the vertex and the x intercept (e3, p244)
x
x
2 2
x
x x
2 6 6
x
6
x
3 3 2 2
x
8 3 2 8 3 1 2 vertex at
x
3 2 8 8 3 ; min/max of function 1 Factor it
x
2
6
x
8
(
x
2 )(
x
4 )
x
2 , 4
x axis intercepts
Given
vertex
and a
point
find the equation of a parabola (e4) (p244) plug in all 4 values given.
y
y
6
a
x a
3
h
2 1 2
k
2
vertex
(
h
,
k
)
a
2 2
x
1 2 2
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
Example 11 (P274): Find the zeros of
f
6
x
3 4
x
2 3
x
2 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) factors 1 factors of of 2 6 1 , 2 1 , 2 , 3 , 6 1 , 2 , 2 , 1 3 , 2 3 , 1 6 Setp3: Test 3 2 6
f
3 6
x
4 2
x
3
x
3 3 3 2 2 0 Plug it into the calculator, don’t try to evaluate it by hand Yes 2/3 is a zero
Setp4: Synthetic Division 2/3 x 3 6 6 4 0 x 2 -4 x 3 0 3 2 0 c -2 6(2/3) Remainder and factor theorems on page 268; e5,6 We will skip the upper and lower bound rule on page 258
f
2 3 6
x
2 3 x-2/3 6x 3 6x 3 6x 2 -4x 2 -4x 2 3 3x -2 3x -2 0 Remainder is 0 Another proof that 2/3 is a zero Now the long division is much easier (p264, e1,2,3)
Example 3 (p28): Step1: Plot it
f
x
5
x
3 2
x
2 12
x
8 Possible zeros, repeated (touches)?
A zero Step2: Rational Zero Test (p 256) factors factors of of 8 1 1 , 2 , 4 , 1 8 1 , 2 , 4 , 8 Setp3: Test: Plug in 1 and 2
x x
2 2
x x
1 1 ( (
x x
1 )
x
2 4
x
1 )(
x
2
i
)(
x
2
i
) 2
x
1 (
x
1 )
x
2 2 Setp4: Synthetic Division x 5 x 4 x 3 x 2 x c -2 1 0 -2 1 4 2 -10 -12 16 8 -8 1 -2 5 -8 4 0
x
2
x
4 2
x
3 5
x
2 8
x
4 1 x 4 1 1 x 3 -2 1 -1 x 2 5 -1 4 x -8 4 -4 c 4 -4 0
x
2
x
1
x
3
x
2 4
x
4 1 x 3 1 x 2 -1 1 x 4 0 c -4 4 1 0 4 0
x
2
x
1 (
x
1 )
x
2 4
Example 1, p286 Rational Functions and Asymptotes (3.5) As x (input) gets smaller y (output) gets larger As x (input) gets bigger y (output) gets smaller
asymptote
a 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 0 domain
f
(
x
) 1
x f f
.
001 1 100 100 .
01
Degree of the numerator is less than the degree of the denominator Horizontal asymptote: y = 0 Degree of the numerator is equal to the degree of the denominator Horizontal asymptote: y= ratio of leading coefficients 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)
Example 3 (p298) Degree of numerator < denominator Horizontal asymptote y = 0 set x to
f
0 0 y intercept 0
set
y to 0
x x
2
x
x
0 2 0
Slant Asymptotes – Page 299
f
x
2
x x
1
x
1
x x
1 2 2
x
2
x
2 1 Slant asmyptote Synthetic Division x 2 x -1 1 -1 1 -1 -2 2 2 c 0
Example 5 (p299)
f
x
2
x x
1 2
x
1
x
1 2
x
x
2 1 Slant asmyptote 1 x 2 1 1 1 0 x -1 c -2 0 -2