Question

what exponential function passes thru (0,4) and (-2,64)

Answer 1

the basic form for an exponential equation is

\[y = ax^2 + b\]
so I'd plug in (0,4) and we can solve for b
then plug in (-2,64) to solve for a

Answer 2

\[4 = 0*0^2 + b\]\[b = 4\]\[64 = a*(-2)^2 + 4\]\[15 = a\]

boom. plug those in your general equation (above) and you've got an answer!

Answer 3

these are the choices tho 1) f(x)=4(1\4)^x
2) f(x)=4^x
3)f(x)=4(1\4)^-x
4)f(x)=-4(1\4)^x

Answer 4

thats why im confused?

Answer 5

well you can plug each set of numbers into all those functions and see which works.

it looks like (1) will work for both sets.

Answer 6

ahh. i misread 'exponential' to be 'quadratic'. My answer is one example, but there are many functions that would work.

Answer 7

thank you so much and by any chance do you know this one: 3 sides of a rectangular garden are enclosed using 120 meters of fencing. what is the maximum area of garden?

Answer 8

hmm what level math are you in? I am tempted to use derivatives, but not if you haven't learned them yet.

Answer 9

algebra 2

Answer 10

and yeah i havent learned that yet lol

Answer 11

well I'm thinking that the maximum area would be a perfect square rather than a rectangle, don't you think? this would mean each of the three sides are equal.

Answer 12

The basic form is\[y=a^{bx}\]

Answer 13

ok we have x+x+y=120 so y=120-2x
so A=xy=x(120-2x)=120x-2x^2 so if we find the vertex of this parabola we can find the max

Answer 14

I get \[y=2^{-3x}\]

Answer 15

i got y=4^(-3x/2) for that one

Answer 16

Actually, I don't...!

Answer 17

so with the choices 1) 3,600m^2 2) 1,800m^2 3)1,600m^2 4)900m^2 whats the correct answer

Answer 18

anyways the vertex of that parabola is (0,120) and you find that by writing the parabole as a*(x-h)^2+k where (h,k) is the vertex

Answer 19

oops one sec i made a mistake

Answer 20

\[y=4e^{-x \ln4}\]

Answer 21

If you let y be length of one of the sides, and x be the length of the other two sides, then y+2x=120. The area is given by A=xy, so since y=120-2x, just substitute for y in A=xy. This gives A=x(120-2x). You are trying to find x that maximizes this. Since the graph of this will be an upside down parabola with roots at x=0 and x=60, this means the maximum will occur at x=30 (halfway between the roots). So the max area is A=30*(120-2*30)=30*60=1800.

Answer 22

When x is 0, y is 4.
When x is -2, y is 64.

Answer 23

the vertex of the parabola is (30,1800) since A=-2(x-30)^2+1800

Answer 24

Why are we talking about quadratics?

Answer 25

oh okay so the answe is 2

Answer 26

1800

Answer 27

\[y=ae^{bx}\]When x=0, y=4, so\[4=ae^0=a\]so a=4.

When x=-2, y=64, so\[64=4e^{-2b} \rightarrow 16=e^{-2b} \rightarrow \ln 16 = -2b \rightarrow b=-\frac{1}{2}\ln 16=-\ln 4\]so b = -ln(4). hence your equation is,\[y=4e^{-(\ln 4)x }\]

Answer 28

you're so smart lokisan lol

Answer 29

haha...but wait, there's more...

Answer 30

\[e^{-(\ln4)x}=e^{\ln \frac{1}{4}x}=\left( e^{\ln \frac{1}{4}} \right)^x=\frac{1}{4^x}\]so\[y=4e^{-(\ln 4)x}=4.\frac{1}{4^x}=\frac{1}{4^{x-1}}=4^{1-x}\]

Answer 31

So your equation can be written as,\[y=4^{1-x}\]

Answer 32

Silence...

Answer 33

lol u guys are so good at this im so confused. but okay if thats my equation whats the answer, 1 2 3 or 4

Answer 34

3600 or 1800 or 1600 or 900

Answer 35

What do you actually need to find out? I thought you just had to come up with the equation.

Answer 36

imrickjames is here...haven't seen you in a while ;]

Answer 37

noo .. i wrote up there earlier the choices...its asking me what the maximum area of the garden is

Answer 38

Okay...let me check the multiple choices...

Answer 39

What garden? Isn't that another question?

Answer 40

Oh, I see the garden question. So the others didn't work it out?

Answer 41

no i just need to answer the garden question. i need to choose one of those 4 and show work.

Answer 42

1800

Answer 43

Thanks for fanning me, Matt :)

Answer 44

thank u ..wat about if a car is purchased for 20,000. the value of the car decreases by 10% each year. about how many years will it take for the vallue of the car to reach 8,000

Answer 45

hey! you can't know it was me! heh :). impressed by your scan.

Answer 46

About 8.7 years.

Answer 47

Impressed by the scan...so *that's* how I increase the fan base ;)

Answer 48

how did u get that thoo

Answer 49

These types of equations with growth and decay in proportion to themselves follow the form\[p(t)=p(0)e^{kt}\]where p(0) is the initial value and k some constant. For growth, it's positive, for decay, negative. You get this equation by solving the differential equation,\[\frac{dp}{dt}=kp\]which is separable (I don't know what level you're at).

Anyway, with this equation, you use your data to find out what p(0) and k are, then answer the question.

Answer 50

Here, p(0) is 20,000.
To find k, you know that after 1 year, the car's value is 10% less, so it's value at time t=1 is 18,000. So,\[18,000=20,000e^{k}\]Divide both sides by 20,000 and take the natural log to find k. Now you have your p(0) and your k, so you can find p(t) at any time t.

Answer 51

okay thank uuu ! and if i have to find |3+i| that equals what ?

Answer 52

You want to find t when p(t) is 8000, so you set it up as,\[8000=20000e^{ \ln (\frac{9}{10})t}\]You have to solve for t. Divide both sides by 20,000 and again, take the natural log to find\[\frac{8000}{20000}=\ln (\frac{9}{10})t \rightarrow t =\frac{8000/20000}{\ln (9/10)} \approx 8.7yrs\]

Answer 53

It's asking for the magnitude. The magnitude of a complex number is z is,\[|z|=|x+iy|=\sqrt{x^2+y^2}\]You consider the real and imaginary part (note, the imaginary part is NOT iy, it's y, and y is real...yi is imaginary).

Answer 54

\[|3+i|=\sqrt{3^2+1^2}=\sqrt{10}\]

Answer 55

Good luck. If you have anymore questions, post them separately...I need to log out :D

Answer 56

thank youuuuu so muchhh