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

- anonymous

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

- katieb

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- mattfeury

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

- mattfeury

\[4 = 0*0^2 + b\]\[b = 4\]\[64 = a*(-2)^2 + 4\]\[15 = a\]
boom. plug those in your general equation (above) and you've got an answer!

- anonymous

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

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## More answers

- anonymous

thats why im confused?

- mattfeury

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.

- mattfeury

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

- anonymous

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?

- mattfeury

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

- anonymous

algebra 2

- anonymous

and yeah i havent learned that yet lol

- mattfeury

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.

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

Actually, I don't...!

- anonymous

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

- myininaya

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

- myininaya

oops one sec i made a mistake

- anonymous

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

- anonymous

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.

- anonymous

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

- myininaya

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

- anonymous

Why are we talking about quadratics?

- anonymous

oh okay so the answe is 2

- anonymous

1800

- anonymous

\[y=ae^{bx}\]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,\[y=4e^{-(\ln 4)x }\]

- myininaya

you're so smart lokisan lol

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Silence...

- anonymous

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

- anonymous

3600 or 1800 or 1600 or 900

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

What garden? Isn't that another question?

- anonymous

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

- anonymous

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

- anonymous

1800

- anonymous

##### 1 Attachment

- anonymous

Thanks for fanning me, Matt :)

- anonymous

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

- mattfeury

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

- anonymous

About 8.7 years.

- anonymous

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

- anonymous

how did u get that thoo

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

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\[\frac{8000}{20000}=\ln (\frac{9}{10})t \rightarrow t =\frac{8000/20000}{\ln (9/10)} \approx 8.7yrs\]

- anonymous

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).

- anonymous

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

- anonymous

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

- anonymous

thank youuuuu so muchhh

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