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Mathematics 14 Online
OpenStudy (anonymous):

GRAPHING EXPONENTIAL FUNCTIONS 1. If (-1, y) lies on the graph of y = 3^x+1, then y = a) 0 b) 1/3 c) 1 2. If (x, 1/100) lies on the graph of y = 10^x, then x = a) 2 b) -1/2 c) -2 3. If (-1, y) lies on the graph of y = 2^2x, then y = a) -4 b) 1/4 c) 1 4. What is the relationship between the graphs of y = 2^x and y = 2^-x? a) reflections over the x-axis b) reflections over the y-axis c) reflections over y = x 5. If (3, y) lies on the graph of y = -(2x), then y = a) 1/8 b) -6 c) -8 6. Which of the following equations is not exponential? a) y = 1x b) y = (1/2)x c) y = -2x 7. If (-2, y) lies on the graph of y = 4^x, then y = a) -16 b) 1/16 c) 16 8. If (-3, y) lies on the graph of y = 3^-x, then y = a) -27 b) 1/27 c) 27 9. If (-3, y) lies on the graph of y = 3^x, then y = a) 1/27 b) -1 c) -27 10. If (-3, y) lies on the graph of y = (1/2)^x, then y = a) 8 b) 1/8 c) -8

OpenStudy (mertsj):

Is the first one \[3^x+1\]

OpenStudy (mertsj):

or\[3^{x+1}\]

OpenStudy (anonymous):

the second

OpenStudy (mertsj):

Did you replace x with -1?

OpenStudy (anonymous):

no

OpenStudy (mertsj):

If you want to find y and y = 3^(x+1) when x = -1 then you should replace x with -1

OpenStudy (anonymous):

so y = 3^(x-1)

OpenStudy (mertsj):

What does the first problem say y is equal to?

OpenStudy (anonymous):

3

OpenStudy (mertsj):

I guess you are reading a different problem than the one you posted. I am so confused. You posted that y = 3^(x+1) and now you are telling me that y = 3

OpenStudy (mertsj):

Supposing you have the following equation:

OpenStudy (mertsj):

|dw:1362515763290:dw|

OpenStudy (mertsj):

And I say to you that the point (3,y) is on the graph. What is the value of x I just gave you?

OpenStudy (mertsj):

Hello? Are you there?

OpenStudy (anonymous):

yes hold up 3?

OpenStudy (anonymous):

are you there

OpenStudy (anonymous):

Hello? Are you there?

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

ok could you figure them out?

OpenStudy (mertsj):

I can figure them out. Can you?

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