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# double check

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5 2.   Dec 12, 2019

#1
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First one

Center  =  (1,4)  = (h, k)

Since the transverse axis is parallelto the y axis, the hyperbola opens up and down

The form is

(y - k)^2            (x  -h)^2

______    -     _________   =    1

a^2                  b^2

a = 20/2  =  10  ⇒  a^2  = 100

b = 16/2 =  8  ⇒ b^2  = 64

So.....the equation is

(y - 4)^2        (x - 1)^2

______  -     _______   =     1

100                64

Here's the graph :

https://www.desmos.com/calculator/ppuyhib2r6   Dec 12, 2019
#2
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ohh okay i see my mistake thank you!!

jjennylove  Dec 12, 2019
#3
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Second  one.....the easiest way to  see  this one is to note what happens to the lead term of each polynomial  as we approach  either a large negative  or a large positive

1.    As  - 2x^3    approaches  some large negative value.....the limit of the polynomial  approaches  positive infinity.....so.....True is  correct!!!

2.  As 2x^4  approcahes some large  positive value......the polynomial  approaches positive infinity......so....True is correct

3.   As  -9x^5  approaches some  large positive value, the polynomial approaches negative infinity....so.... True is correct   Dec 12, 2019
#4
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Third one

(x + 2)^2           ( y + 8)^2

_______    +    _________  =     1

16                     81

Center   (-2, -8)

The major  axis  =   2sqrt (81)  =  2 * 9  =   18  units long....so....the enpoints of the major axis are  18/2 =  9 units from the center

The minor  axis =  2sqrt(16)  = 2 * 4  = 8  units long....so....the endpoints of the minor axis  are  8/2  = 4 units from the center

The endpoints  are  ( -2  + 4 , -8)  ( -2 - 4, -8)   ( - 2,  -8 + 9)    (-2, -8 - 9)   =

(2, -8)   ( -6, -8)  (-2, 1)   (-2 - 17)

Connect these with a smooth curve to graph the ellipse....here it is  :

https://www.desmos.com/calculator/jwlnfr0asu   Dec 12, 2019
#5
+1   