Rational Functions
What is the vertical asymptote of this function?
f(x) = (1/(x+4)) +5
x=-4
(x2+2)-(x2+1)
12^400=12^x+48
x=?
x=400-48= 352
x=352
limx-->9(3x2+4x+2)
281
What is the horizontal asymptote of this function?
f(x) = (x2+3x-9)/(4x-4x2+2)
y=-1/4
(13x3+3x2-x+5)+(5x3-x2+x+3)
18x3+2x2+8
Rewrite in logarithmic form 9^-2=1/81
log9(1/81)=-2
Find the center and major vertices of the following ellipse:
(x+6)2/49 + (y-3)2/81 = 1
center: (-6,3)
major vertices: (-6,-6) (-6,12)
Is there a hole (jump discontinuity) in this function? If so, where (at what value of f(x))?
f(x)=(x2-x-12)(x+1)/(x2-4x-21)
When x=-3.
f(x) is -1.4
(x3-2x2+5x-3)(3x-1)
3x4-7x3+17x2-14x+3
You purchased a computer for $1200 and in 7 years it’s now $500 what is the depreciation rate of the computer?
500=1200(1-7x)
(1-7x)=500/1200
5/12=1-7x
-7x=(5/12)-1
-7=-7/12
x=-7/12/-7
x=0.0833
limx-->10(x+3)/(x-10)
DNE
What is the slope of this asymptote...(Hint: It is a slant asymptote).
f(x)=(x3+4x2-6x+36)/(x2-4)
SA: y=x+4
Solve for x:
3x3+9x2+3x+9=0
x=1, -1, -3
$10000 is invested at a rate of 4% compounded quarterly. How long will it take to end up with $30000 in the bank?
C=P((1+.04)^4n -1)
30000=10000(1.04)^4n -1
3=(1.04)^4n - 1
-4 = -(26/25)^4n
4n=2log1.04(2)
n=8.84
Find the limith-->0 (f(x+h)-f(x))/h given
f(x)=(sqrt x+4)
1/(2(sqrt x+4))
There is a jump discontinuity (hole) in this function. Find it, and find the value of f(x) at this point.
f(x)=(x4+5x3-2x2-24x)/(x2-16)
when x=-4
value of f(x) is 3
(2x3+4x2+x+1)/(x+1)
2x2+1
If you invest $4000 with a weekly compound of a 6% interest rate how much will you have in 15 years
A=P(1+r/n)^nt
A=4000(1+.06/48)^52*15
A=4000(1.00125)^780
A=$10,598.21
Find the center, vertices, and foci points for the following hyperbola:
x2-9y2+4x+108y-356=0
center: (-2,6)
major vertices: (-8,6) (4,6)
minor vertices: (-2,8) (-2,4)
foci: (-2-2(sqrt 10),6) (-2+2(sqrt 10),6)