log b (M) = log b (N)
log b (M) - log b (N)
K * log b (M)
0
1
K
K
Log c (b)/ log c (A)
x/y
1/y * log a (b)
P(t) = Po (2) ^ t/d
A(t) = Ao(1/2) ^ t/n
replace all x values with y, tov
Reflect diagonally
Y = a log b (k (x-d)) + c
VS or VR = bafo a
HS or HR = bafo 1/k
Ht = d, positive to the right, negative to the left
Vt = c
b^c=a
x values
y values
(F o g)(x)
sub in other function as x values
reverse the opperation done
F(x) = ?
F(x+1)=x^2+ 2x+1
F(x) = (x-1)^2 + 2(x-1) + 1
inverse functions
1/f(x)
points
o = undefinded - asymptote
(-+) 1 = invarient
less than 0 = less than 0
more than 0 = more than 0
curvy!
max/min becomes 1/max or min
ax^2 + bx + c
sign of a determines opening direction
x = - b/2a is axis of symmetry
Y= a (x-h)^2 + k
k is the max or min, the vertex point
h is the x coordinate of the vertex
Y=a (x-q) (x-p)
gets you the opposite of the intercepts :)
vertex to standard we complete the square
-x^2+6x+7
x^2 + 6x + (6/2)^2 + 7 - (6/2)^2
(x+3)^2 +7-9
(x+3)^2 -2
Un = U1 + (n-1) * d
d being the common difference
d= change in Un/ change in n
Sn= n/2 (2U1 + (n-1)*d)
n is the end term,
uk is the formula
k indicates where to start
Un = u1 r ^n-1
r being the common ratio
Sn = U1 (1-r^n) / 1-r
S infinity = u1/ 1-r
A = P (1 + (r/n)) ^ (n*t)
intail money ( one plus intrest over compounding periods in a year) to the power of compounding periods in a year divided by the amount of years
A = P (1 + nr)
initial money times 1 + (compounding period time intrest rate)
Use discriminant
triangle = b^2 - 4ac
more than one = two real roots
one = one root
less than one = imaginary roots, no real roots