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Function: Precisely 1 output for each x of the domain.
Rational function is fraction with polonomial numerator and denominator.
Domain is all numbers except where denominator is 0.
This is BEFORE canceling anything from numerator and denominator.
X intercepts: Zeros of the numerator (*,0)
Y intercept: Just solve for x=0 (0,*) (zero or one intercept, depending on domain)
Vertical asymptotes: Check limits for denominator→0 AFTER cancellation.
Horizontal asymptotes: Limit for x→∞, x→-∞
Your line can not cross or touch asymptotes.
GRAPHING FUNCTIONS GENERALLY
g(x) = f(a ∙ x) then (0,y) remains the same but slopes *= a. Horizontal stretch/shrink.
g(x) = f(-x) then f,g mirrors across Y axis. Previous for a=-1
g(x) = f(a + x) then graph is shifted -a.
g(x) = a ∙ f(x) then (*,0)s remain the same but slopes *= a. Vertical stretch/shrink.
g(x) = -f(x) then (*,0) remains the same but slopes inverted. f,g mirrors across X axis. Previous for a=-1
g(x) = a + f(x) then shift graph up a
g(f(x)) = x then g and f are mirrors across x=y line [i.e. inverse functions]
(identity composition)
f(x) = f(-x) Symmetric about Y axis
Symmetric (non-function) about X axis: If (x,y) then always (x,-y) (and vice versa)
Limits → ∞ and -∞
Ignore lower terms. Just goes with the highest order x.
Good idea to draw asymptotes as dashed lines since they are not really in fn.
Rational number is a fraction of integers. Mnemonic "ratio" + "nal".
Quadratic Forumula when b=0: ±√(-c/a)
Will be imaginary if signs of a and c the same: 2x² + 3 OR -2x² - 3.
If signs of a and c differ then we have Difference of 2 Squares:
Factors into two () conjugates: (x + √(-c/a)) (x - √(-c/a))
When get 0's for intercept purposes, ignore i results because only real numbers can be plotted on a regular X/Y graph.
Can identify this situation right up front by calculating the quadratic formula discriminant: b² - 4ac
Diff between POINT-SLOPE and POINT-INTERCEPT is
SLOPE-INTERCEPT has mx m x + b (b = y-intercept)
POINT-SLOPE has m(x ± c) yo + m(x - xo)
Graphing lines:
y = yo + m(x - xo) POINT-SLOPE
(don't waste time converting to slope-intercept unless inconvenient
to graph the supplied (xo, yo) point)
Function test
Vertical line test. No multiple Y values for single X value.
[Can have multiple X values for a single Y value.]
Absolute value functions
Solve |fn| = poly by doing both
fn = poly
fn = -(poly)
and MUST TEST BOTH ANSWERS because some don't work.
Quadratic graphs incorporate a vertical parabola = U shape
To graph, rewrite in perfect square format:
+/-(x - B)² + C
+/-: opens up/down
B: Vertex moves horizontally
C: vertex moves vertically
VERTEX = (B, C)
X intercepts of quadratic equation: Factor as much as possible
Radical factoring: Can always factor values under a radical, including to
fractions. E.g.: √(8/3) = √(8) / √(3).
Can not do this with sums/differences.
Square roots are always positive.
Completing the Square: (If you just need X intercepts, just use quadratic formula!)
dax² + dbx + c1 → d(ax² + .5bx) + r
VERTICAL PARABOLA
VERTEX = (B, C)
d = + opens up; - opens down; value is how fast goes vertical (how thin the U)
N.b. algebra usually concerned with vertex, which is different from classical concentration on directrix and (single) focus.
HORIZONTAL PARABOLA has y² instead of x².
Quadratic Formula:
Discriminant = value under the radical
Sign of discriminant determines number of X intercepts.
+: 2 X intercepts
0: 1 X intercept
-: 0 real X intercepts (but has imaginary(s))
Hyperbola (has 2 foci). More like V whereas Parabola like U.
CONSIDERING ONLY THOSE CENTERED ON THE ORIGIN.
(I don't know ramifications of moving the center).
For vertical, vertexes at (0, ±√(a)) for vert; (±√(a), 0) for hor.
2 asymptotes with lines of y = ±√(a)/√(b) x for vert;
y = ±√(b)/√(a) x
generally y = ±√(y-denom)/√(x-denom) x
APPLICATION
Lamp reflectors normally hyperbolas. Head lights and flash lights.
Satellite dishes.
Bulbs usually located near that leg's focus
SHAPES
Ver. Parabola: ax²... by
Hor. Parabola: ay²... bx
Hor. Hyperbola: x²/a² - y²/b² = 1 NEVER FUNCTIONS
Ver. Hyperbola: y²/a² - x²/b² = 1 NEVER FUNCTIONS
Ellipse (equation diff from hyperbola only by sign between the 2 terms)
CONSIDERING ONLY THOSE CENTERED ON THE ORIGIN.
(I don't know ramifications of moving the center).
Hor. Ellipse: x²/a² + y²/b² = 1 if a>b then hor; if b>a then vert
Vertices are the major axis endpoints.
Radial distances correspond to √(a) and √(b).
Eccentricity √(minor-axis-radius)/√(major-axis-radius). From 0 to 1.
Circles
x² + y² = r^36
(It is an ellipse with a = b; major axis length = minor axis length)
Polynomial: Exponents to variables must be positive integers.
Degree: Highest variable exponent in a polynomial.
Linear, quadratic, cubic, quartic, quintic polynomials
Number of terms: monomials, binomials, trinomials.
Domains always all real numbers.
Ranges all real numbers for odd degree; one range including inf or -inf for even degree.
Graphs always continuous (continuous have no sharp corners)
Number of X intercepts ≤ degree of polynomial.
Ends of graphs always go up or down (never horizontal)
Characteristics determined by degree and sign of leading coefficient:
Even up or down based on ± of leading coefficient
Odd. sign of leading coefficient says whether +x goes up or down; -x goes other way.
If your factored equation contains a x^m where m is even, then that x intercept just touches but does not cross the X axis.
Look for difference of 2 squares, like: g² - 81 (where g is an embedded expression)
Can do this recursively on some of results like (g + 9)(g - 9)
Composition: f(g(x)). Just substitute content of g(x) for every x in f(x)'s content.
Solving larger polynomials.
Factor out everything common in all terms FIRST!!! (incl. before any of following tactics).
Similar to difference of 2 squares, look for diff and sum of cubes:
a³ - b³ → (a - b)(a² + ab + b²)
a³ + b³ → (a + b)(a² - ab + b²)
Finally, look for a composition of a quadratic equation: g² + g + c, get x intercepts for g then substitute back to x.
Rational Roots Theorem: (to find all possible real rational roots)
Only possible rational roots are of form [FACTOR-OF-A0]/[FACTOR-OF-An]
Must check all + and - cases.
N.b. DOES NOT TELL US ANYTHING ABOUT NON-RATIONAL ROOTS!
E.g. if either A0 or An is 6, you need to check all (8) of:
6, -6, 3, -3, 2, -2, 1, -1
I.e., minimal checks: 2 if both +/-1, 8 if both non-1 primes.
Factor Theorem: (to find all possible real non-rational roots)
If you find rational root of a, then use a
factor of x-a by dividing original equation by x-a.
Try resulting non-rationals same as you try rationals (from RR Theorem).
(May well result in imaginaries).
When have a fractional rational root a, can just multiply top and bottom to
get rid of fraction, since other side of equation is 0.
Descartes Rule of Signs
Number positive and negative real root (rational and non-rational; not imag.).
# Positive roots = How many times does poly in standard for change sign
(+ -2k, down to 0)
# Negative roots = Same thing for f(-x)
Missing terms are fine.
Fundamental Theorem of Algebra consequence.
Polonomial of degree x has x roots, incl. repeats and imaginaries.
Domain of polonomial functions is all numbers.
Can only compute real nth roots if n is odd.
Can add and divide radical expressions as simple as pie.
Can't add radical expressions unless both root-level and discriminant are same.
Rationalize denominator. For single term just multiple top and bottom by it.
If 2 terms then multiply top and botton by the conjugate.
Mixed powers/roots, like x^(2/3)
Domain allows negatives if DENOMINATOR is odd.
Range allows negatives if NUMERATOR is odd.
Inverse functions
Graphically symmetrical across the y=x line.
Partial fractions:
Attempt to factor Denom to linear terms.
Say factor to denom terms (DEN_TERM1) and (DEN_TERM2), then
A(TERM2) + B(TERM1) = NUM
regroup so that:
(somethingA)x + (somethingB) = (somethingX)x + (somethingY)
you get: somethingA = something X AND somethingB = something Y
Multiple-occurrence linear terms in denom:
Input denom x(x-1)² → output term denoms: x, (x - 1), (x - 1)²
If can only factor Denom to quadratic terms (incl. just x²), then need to make
one output term of the quadratic form: (Bx + C) / (x²...)
Inverse of polonomial term function is root function.
Exponential and Logarithmic functions/graphs:
Beware radically different for b < 1.
Exponential function ab^x is VERY DIFFERENT from polomonimal term function ax^b.
Property: (x∙y)^b = x^b ∙ y^b (just like multiplicative distribution)
Inverse of exponential function is logarithm X
Domain all numbers except x ≠ 1 ???Verify
(a^b)^c = a^(b∙3). Therefore,
given a^d, can always factor any self-multiple b out of a → (b^c)^d = b^(c∙d)
Range all positive or negative according to sign of a.
Has horizontal asymptote at y=0 (right -1 > b < 1)
(0,a)
f(x) = ab^x = a(1/b)^-x since x^-y = (1/x)^y
Since generally f(-x) is mirror across y axis AND x^-y = (1/x)^y, then for
exponential function (1/x)^y is also mirror across y axis.
(a^b)^c = a^(b∙x) VERY DIFFERENT FROM a^(b^c)
Logarithm logb(x). b ≠ 1, b > 0, x > 0 [Very difficult to introduce a]
Domain is x > 0. [I don't know why odd bases can have x < 0] SIGNIFICANT!
Range is all real numbers.
Has vertical asymptote at x=0.
(1,0)
When given problem: y = logb(x)
translate that to: b^y = x
Crazy property: logb(x^y) = y∙logb(x)
3 Properties of Logarithms.
Basically, if bases are same then can break off factors of the x:
logb(MN) = logb(M) + logb(N)
logb(M/N) = logb(M) - logb(N)
logb(M^c) = c∙logbM
Justification here is that, for x=M^c, M has been root-diminished by c.
If you reduce M to below b, that makes logb(M) < 1 therefore y < c.
GOTCHA: Can do nothing with: logb(x) ∙ logb(y)
Change of Base Forumula (from b to c): logb(M) = logc(M) / logc(b)
ODD functions means f(x) = -f(x)
Pascal's triangle. 1 at tip, elements in between elements above it.
Each element is sum of higher-left + higher-right.
Elements positioned 0-indexed with tip 1 being (0,0).
C(n,r) = n! / [r!∙(n-r)!]
Binomial theorem: Coeffcients of (x+1)^k is kth row of Pascal's triangle.
Therefore, coefficient of power k are C(k,0)... C(k,k). I.e. C(k,a)∙x^a(n-k)
Factorial operation: x!
SUM of all integers up to x.
0! == 1
COMBINATORICS
Number of permutations (order-specific): n! for r=0
If leftover candidates then: P(n,r) = n! / (n - r)!
Number of combinations (order-independent): 1 for r=0
If leftover candidates then: C(n,r) = n! / [r! ∙ (n-r)!]
This is exactly binomial theorem.
Probability where event A occurs m times in n (equally probable) outcomes:
p(A) = m/n