Question 302288: Can someone please help me with this problem? please. Evaluate the exponential equation for three positive values of x, three negative values of x, and at x=0. Show your work. Use the resulting ordered pairs to plot the graph; submit the graph via the Dropbox. State the equation of the line asymptotic to the graph (if any) Y = 1/4^x+3
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! It's similar to this one:
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evaluate the exponential equation for three values of x greater than -4, three values of x smaller than -4, and at x=-4. use resulting pairs to plot the graph. state the equation of the line asymptotic to the graph. please help me, I am trying really hard to understand this but don't think I get it. y=3^x+4 +1 the pairs I have got are (-3,4) (-2,10) (-1,28) (-4,2) this is as far as I can get. And don't really understand how to do the graph. any help would be greatly appreciated.
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Do the same for additional points:
x....y
0....2
-5...4/3
-6...10/9
-7...28/27
-8...82/81
You can use Excel to find additional points.
To graph is, connect the points with a curve. Add points as necessary, eg, between 10 and 28.
dl FREE graph software from
http://www.padowan.dk.com
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website!
Y = (1/4)^x+3
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Evaluate the exponential equation for three positive values of x, three negative values of x, and at x=0. Show your work.
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x = -3 ; y = (1/4)^(-3)+3 = 4^3+3 69:::::(-3,69)
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x = -2 ; y = (1/4)^(-2)+3 = 4^2+3 = 19:::(-2,19)
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x = -1 ; y = (1/4)^(-1)+3 = 4 + 3 = 7::: (-1,7)
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x = 0 ; y = (1/4)^0+3 = 1+3 = 4::::::::: (0,4)
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x = 1 ; y = (1/4)^1+3 = 3 1/4::::::::::: (1, 3 1/4)
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x = 2 ; y = (1/4)^2+3 = 3 1/16 ::::::::: (2, 3 1/16)
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x = 3 ; y = (1/4)^3+3 = 3 1/64 ::::::::: (3,3 1/64)
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Cheers,
Stan H.
Use the resulting ordered pairs to plot the graph; submit the graph via the Dropbox. State the equation of the line asymptotic to the graph (if any) Y = 1/4^x+3
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