our last reflection!! okay anyway, y'all here are some stuff from chapters 1 - 6 & 13..im gonna print it out and use it to study..i hope y'all will do the same :) good luck
Completing the Square:
You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².
For example:
x² + 6x - 2 = 0
* anytime you are solving a quadratic you’re finding x-intercepts
* Move the constant term to the right side:
x² + 6x = 2
* Take half of the coefficient on the x-term (divide it by two, and keeping the sign), and then square it. Add the squared value to both sides of the equation:
x² + 6x + 9 = -2 + 9
* Convert the left-hand side to squared form. Simplify the right-hand side:
(x + 3)² = 7
* the # half of the coefficient goes in the parentheses.
* Square-root both sides:
x + 3 = √7
* Solve for "x =". Remember to put the "±" on the right side and that it gives you two solutions.
x = -3 ± √7
* The two points for this solution are:
(-3 + √7,0) , (-3 -√7,0)
Rational Root therom
Example: f(x)= 2x^3 + 3x^2 - 8 + 3
Step 1: find all possible roots..
p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2
*p is the leading constant term & q is the leading coefficient
possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2
Step 2: now you can plug all of the possible roots in your calculator to find the roots that work
* the zero will be: 1, 1/2, -3
Step 3: use synthetic division to factor all of the roots that work
you should get: (x - 1) (2x^2 + 5x + 3)
Domain & Range of functions:
Polynomials:
the domain of all polynomials is (−∞, ∞).
For example:
f(x) = x^2 - 3x^2 + 2x - 1
D: (−∞,∞ )
f(x) = x^2 + 3
D:(−∞,∞ )
Fraction:
* you set the bottom to zero
* solve for x
* then set up intervals
For example:
f(x) = 1/x-2
x-2=0
x=2
D: (-∞ , 2) (2, ∞ )
Absolute Value:
D:(- ∞ , + ∞)
R: [0 , + ∞)
For example:
f(x) = |x + 8| - 9
D: (- ∞ , + ∞)
R: (-9, ∞)
f(x) = |x -7| + 5
D: (- ∞ , + ∞)
R: (5, ∞)
Square Roots:
to find the domain:
* set the inside = to zero
* then set a # line
* try values on either side of each #
* get ride of the negatives
* set up intervals
to find the range:
* graph
For example:
√9 - x^2
(solve for x...)
9 - x^2 = 0
-x ^2 = -9
√x^2 = √9
x = ±3
(# line)
(#s on either side..)
f(-4) = √9 - (-4)^2 = √-7
f(0) = √9 - (-4)^2 = √9
f(x) = √4 - (-4)^2 = √-7
√9 = ±3 so...
D: [-3, 3]
(graph....)
R: [0, 3]
How to Find the Inverse of a Function:
* Replace f(x) with y
* Reverse the roles of x and y
* Solve for y in terms of x
* Replace y with f-1(x)
Example 1 - f(x) = 2x + 3
1. write the function as an equation: y = 2x + 3
2. solve for x: x = (y - 3)/2
3. now write f-1(y) as follows .
f -1(y) = (y - 3)/2 or f -1(x) = (x - 3)/2
4. Check:
* f(f -1(x))=2(f -1(x)) + 3
=2((x-3)/2)+3 =(x-3)+3 =x
* f -1(f(x))=f -1(2x+3)
=((2x+3)-3)/2 =2x/2 =x
Example 2 - f(x) = √x + 4
1. (x)^2 = (√y + 4)^2
2. x^2 = y + 4
3. y = x^2 - 4
4. f-1(x) = (x^2 - 4)
* f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x
* f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x
Exponents:
1. b^x * b^y = b^x + y....example: 2^3 * 2^5 = 2^8
2. b^x/b^y = b^x - y....example: 5^7/5^4 = 5^3
3. (ab)^x = a^xb^x....example: (3 * 7)^3 = 3^3 * 7^3
4. (a/b)^x = a^x/b^x....example: (3/5)^3 = 3^3/5^3
5. (b^x)^y = b^xy....example: (2^2)^3 = 2^6
6. b^x/y = y^√b^x....examples: 5^3/4 = 3^√5^3
7. to solve for exponents:
* write as the same base
* set exponents equal
* then solve for x
here are some examples:
(a). 5^3x = 5^7x - 2
In this first part we have the same base on both exponentials so there really isn’t much to do other than to set the two exponents equal to each other and solve for x.
3x = 7x - 2
2 = 4x
x = 1/2
So, if we were to plug x = 1/2 into the equation then we would get the same number on both sides of the equal sign.
(b). 4^t^2 = 4^6 - t
t^2 = 6 - t
t^2 - t - 6 = 0
(t - 2) (t + 3) = 0
t = -3, t = 2
In this case we get two solutions to the equation. That is perfectly acceptable so don’t worry about it when it happens.
(c). 3^z = 9^z + 5
Now, in this case we don’t have the same base so we can’t just set exponents equal. However, with a little manipulation of the right side we can get the same base on both exponents. To do this all we need to notice is that 9 = 3^2. Here’s what we get when we use this fact:
3^z = (3^2)^z + 5
Now, we still can’t just set exponents equal since the right side now has two exponents.
3^z = 3^2(z + 5)
We now have the same base and a single exponent on each base so we now set the exponents equal. Doing this gives us....
z = 2(z + 5)
z = 2z + 10
-10 = z
...a solution of z = -10.
Step 4: solve further
(this can be factored...)
= (x - 1) (2x^2 + 5x + 3)
= (x - 1) (2x - 1) (x + 3)
(set x = 0 )
x = 1, 1/2, -3
Logarithm Properties:
* logb MN = logb M + logb N
* logb M/N = logb M - logb N
* logb M^K = K logb M
* logb b^k = k (this one i don't get..maybe i copied it wrong)
* b^logb^k = k
Here are some examples:
1. log 2 + log 3 + log 4 = log 24 (mulitply: 2 x 3 x 4)
2. log 8 + log 5 - log 4 = log 10 (mulitply: 8 x 5 then divide: 40/4)
3. 2 ln 6 - ln 3 = ln 12 (raise 6 to the 2nd power = 36 the divided by 3 = 12)
4. log M - 3 log N = log M/ N^3
5. ln 2 + ln 6 - 1/2 ln 9 = ln 12/3 = ln 4
6. Expand logb MN^2....logb M + 2 logb N
7. Condense log 45 - 2 log 3....log (45/9) = log 5
8. Rewrite in exponetial form: log36 6 = 1/2....36^1/2 = 6
9. Rewrite in logarithmic form: 2^2 = 4....log2 4 = 2
Changing Bases: (Done when you can't solve a log)
* Rewrite it as an exponential
* Take the log of both sides
* Move the variable to the front
* then solve
(use the same steps when solving for x as an exponent when you can't write them as the same base)
examples:
1. log5 10 = x
5^x = 10
log 5^x = log 10
x log 5 = 1
x = 1/log 5
2. 2^x = 7
log 2^x = log 7
x log 2 = log 7
x = log 7/log 2
(remember b-rob might use random symbol so don't panic)
Conics
Ellipses
Steps:
1. find the center
2. determine the major axis
3. find the vertex (± √big denom)
4. find the other intercept ( ± √small denom)
5. find the focus (c^2 = a^2 + b^2)
6. determine the length of the major axis (2√big denom)
7. find the length of the minor axis (2√small denom)
8. finally graph
Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.
x^2/4 + y^2/9 = 1
This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.
Since a^2 = 9 then a = 3 & Since b^2 = 4 then b = 2
Major intercepts: (0, 3), (0, –3)
Length of major axis: 2 √9 = 6
Minor intercepts: (2, 0), (–2, 0)
Length of minor axis: 2√4 = 4
c^2 = a^2 + b^2
= 9 - 4
= 5
= √5
Foci: (0, √5) , (0, -√5)
then you graph your points..
Parabolas:
how to find the axis of symmetry, vertex, focus, & directrx??
1.) to find the axis of symmetry: x = -b/2a
2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:
y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex
3.) to find the focus: 1/4p= the coefficient of x^2 and then add p
Note:
*If opens up, add to y value from vertex, if opens down, subtract
*If opens right, add to x value to vertex, if opens left, subtract)
4.) directrix: is p units behind the vertex
Note:
*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value
Example: x^2 + 1
~vertex:
x = -b/2a
x = 0/2(1) = 0
0^2 + 1 = 1
(0,1)
~Focus:
1/4p = 1
4p = 1
p = 1/4
(0, 1 + 1/4)
(0, 5/4)
~directrix:
y = 1 - 1/4
y = 3/4
CIRCLES
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)
If the equation is not in standard form, you must complete the square to put it in standard form.
If you are given a center and a point, you can use the distance formula to find the radius.
To find the intersection of a line and a circle:
1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.
***Reminder. If your x value is imaginary, then there is no point of intersection.
EX: find the center and radius.
(x-3)^2+(y+7)^2=19 c:(h,k)
center: (3,-7) radius: square root of 19
EX: find the eqn of the circle with the center (1,4) through (3,7)
in the problem you are given a center and a point so you would plug into the distance formula.
square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13. **13 has no root.
Your answer should be (x-1)^2+(y-4)^2=13
Hyperbola
(x^2-h/a^2)+(y^2-k/b^2)=1
* Your center is (h,k)
* your major axis has the larger denominator
13-1
1. sequence-list of numbers
2. two main types: 1). arithmetic-add or subtract 2).geometric-multiply
Formulas:
1. arithmetic-used to find a term: tn . t1 + (n-1)d
**n=term #, t1=first term, d=what you add, tn=term #
2. geometric: tn=t1 . r^(n-1)
**r=what you multiply by..
Examples:
1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...
tn = 3 + (n-1) (2)
tn = 3 +2n - 2
tn = 1 + 2
2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..
**divide the 2nd term by the 1st term to find r
4.5/3 = 3/2 = r
tn = 3 . (3/2)^(n-1)
13-2
Formula for a sequence that involves the previous term: (an - 1)
Examples:
1. Find the recursive definition of: 81, 27, 9,3,...
an = an - 1/3
2. 1, 2, 6, 24, 120, 720, ....
n = 1: 1
n= 2: 2
n = 3: 6
an = n . an - 1
13 -3
Series-List of added or subtracted numbers
**Leave it as a list: do NOT add
Formulas:
1. Arithmetic: Sn = n(t1 + t2)/2
**Finds the sum of the first n terms
2. Geometric: Sn = t1 (1 -r^n)/1-r
Examples:
1. Find the sum of the first 25 terms of the series: 11 + 14 + 17 + 20 + ....
Sn = n (t1 + tn)/2
t25 = 11 + (24)(3)
Sn = 25 (11 + 83)/2
= 1175
2. Find the sum of the first 10 terms of the series: 2-6 + 18 - 54 +...
**This is a geometric sequence and that is because you have to add or subtract the same number for it to be an arithmetic sequence, got it??
r = -6/2 = -3
Sn = t1(1 - r^n)/1-r
= 2(1 -(-3)^10)/1 - (-3)
= 2(-59048)/2
= -29524
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment