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Kimmiey
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horizontal asymptote
 bottom heavy y=0
 eaual y top divided by bottom  so x^{2}/x^{2 } then y=1
 2x^{2}/x^{2 }then y=2

verticle asymptot
 cannot be crossed
 let the numerator equal 0

point discontinuity
a hole in the graph occurs when x=a gives 0/0

slant asymptoe
 top heavy
 divide numerator by denominator

odd function
 symmetry with respect to the origin
 test if f(x)=f(x) the f(x) is odd

symmetry tests
 x axis make y negative and see if it equals out to the original
 y axis make x negative and see if it equals out to the original
 origin make both x and y negative and see if it equals out to the original

points always on the graph of y=a^{x}
(0,1) (1,a) (1,a)

points always on the graph of y=log_{a}(x)
(1,0) (a,1) (1/a, 1)

how to find vertex
is the x value and insert this for x to get y

completing the square
y=2x^{2}+4xc
 2x^{2}+4x c
 2(x^{2}+2x )7
 insert half of middle term
 2(x^{2}+2x+1 )7
 take it out at the end
 2(x^{2}+2x+1 )7 1
 y=a(xh)^{2}+k
 y=2(x+1)^{2}8




