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qwelly [4]
4 years ago
10

On a coordinate plane, an exponential function has a horizontal asymptote of y = 0. The function curves up into the first quadra

nt. It goes through (0, 3), (0.5, 5), (1, 10), (1.5, 23), (2, 50). The given graph represents the function f(x) = 2(5)x. How will the appearance of the graph change if the a value in the function is decreased, but remains greater than 0? The graph will increase at a slower rate. The graph will show a decreasing, rather than increasing, function. The graph will show an initial value that is lower on the y-axis. The graph will increase at a constant additive rate, rather than a multiplicative rate.
Mathematics
2 answers:
aleksandr82 [10.1K]4 years ago
8 0

Answer:

The graph will show an initial value that is lower on the y-axis

Step-by-step explanation:

The exponential function has the next general form:

y = ab^x

where <em>a</em> is the initial amount and <em>b</em> is the base.

If the <em>a</em> value in the function is decreased, but remains greater than 0, the y-intercept of the curve decrease.

ss7ja [257]4 years ago
7 0

Answer: The graph will show an initial value that is lower on the y-axis

Step-by-step explanation: i got it right on edge!

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-10-|1+4b|=-19<br> Y = 0
alexandr402 [8]

Answer:

b=2 or b=-2.5

Step-by-step explanation:

We solve absolute value equations the same as any other equation using inverse operations. However, the inverse operation for absolute value requires one special condition. Since absolute values change a number to a positive regardless if it has a negative sign, we have two possible answers. When we have the absolute value operation isolated, we will set it equal to both the negative and positive option.  

-10-|1+4b|=-19\\-10+10-|1+4b|=-19+10\\-|1+4b|=-9\\\frac{-|1+4b|}{-1} =\frac{-9}{-1} \\|1+4b|=9

We now have the absolute value isolated. We will set it equal to both 9 and -9.

1+4b=9\\1-1+4b=9-1\\4b=8\\b=2

and

1+4b=-9\\1-1+4b=-9-1\\4b=-10\\b=\frac{-10}{4}=\frac{-5}{2}=  -2.5

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4 years ago
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