cts steps
each side is (x-(b/2)) then find remainder, set factor with remainder equal to zero and solve
even degree means there myst be
turning points
degree can be
same degree as minimum or mote than bec ause of imag zeros
number of tp
one less than degree
vertex form
f(x)=a(x-h)^2+k
a>0
point up and vice versa
aos in vertec
x=h, x is coordinate of vertex
from standard to factor form
just factpr
finding x cord of vertex
add zeros together and divide by two using number line
finding y cord of vertex
input x cords to equation and solce
finding aos number line
use middle value then count from middle with divided values
given a vertex use
vertex form and solce for a
goal of cts
from standard to vertex
a>1 cts
use a amount of diagrams and split x evenly among, the use 1/2 of those for length and other 1/2 of those for width then c units to cts
x cord of vertex in quadratuc
-b/2a
multiplicity
the number of times a factor occurs
k odd vs even
cross vs bounce
to detetmine the a value on a graph
look at y int and input zero for x and solve to make equation true using a
odd mult can be
flat
turning p
less than or equal to degree minus 1
even degree must be
parabola with odd amount of turning points
dividing polynomial
dividend is p(x) divisor is d(x) quotient is q(x) when p(x)/d(x)=q(x)+remainder/d(x)
fundamental theorum of algebra
a polynomial has as many roots as its highest degree
complex number
a+bi where a and b are real numbers and i is imaginary
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