SOLUTION: i need help solving this please... i have tried everything i think of but i am stuck the question is: write an exponential function whose graph passes through the points (0,-3) and

Question 883723: i need help solving this please... i have tried everything i think of but i am stuck the question is: write an exponential function whose graph passes through the points (0,-3) and (4,-48). thanks so much your help is very much appreciated.
Found 2 solutions by jim_thompson5910, DrBeeee:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First plug in (x,y) = (0,-3)

y = a*b^x
-3 = a*b^0
-3 = a*1
-3 = a
a = -3

Now plug in a = -3, and (x,y) = (4,-48)

y = a*b^x
y = -3*b^x
-48 = -3*b^4
-48/(-3) = b^4
16 = b^4
b^4 = 16
b = 16^(1/4)
b = 2

So a = -3, b = 2 giving us this

y = a*b^x
y = -3*2^x

So the final answer is
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
An exponential equation is given by
(1) y = a*(e^(bx))
Our job is to find the values of the constants a and b, given that the curve passes through two points of (x,y).
All we do is to form two equations based on the two points and solve them simultaneously for a and b.
First put (x,y) = (0,-3) into (1) to get
(2) -3 = a*(e^(b*0)) or
(3) -3 = a*(e^(0)) or since e^0 = 1 we have
(4) a = -3
Now we have
(5) y = -3*(e^(bx))
Now put the second point (4,-48) into (5) and get
(6) -48 = -3*(e^(4b)) or
(7) (e^(4b)) = (-48)/(-3) or
(8) e^(4b) = 16
Now take the ln of both sides of (8) to get
(9) 4bln(e) = ln(16) or since ln(e) = 1 we have
(10) 4b = ln(16) or
(11) b = ln(16)/4 or
(12) b = ln(2) giving
(13) y = -3*(e^((ln2)x)) which simplifies to
(14) y = -3*(2^x) because e^ln2 = 2
Answer: y = -3*(2^x)
Check your answer. Is y = -48 when x = 4?
Is (-48 = -3*2^4))?
Is (-48 = -3*16)?
Is (-48 = -48)? Yes

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