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ch4aika [34]
4 years ago
14

What is the range of the exponential function shown below? F(x) = 11 • (1/3)x

Mathematics
1 answer:
vlabodo [156]4 years ago
3 0

Answer:

(0, \infty)

Step-by-step explanation:

By definition all the exponential functions of the form

f (x) = a (b) ^ x

Where a is the main coefficient and b is the base they have range (0, \infty)

Whenever a> 0 and b> 0.

In this case the function is:

f (x) = 11(\frac{1}{3}) ^ x

Note that for this function a = 11> 0 and b =\frac{1}{3}>0

Therefor the range is: (0, \infty)

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Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

  • m is the slope
  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

  • (-3,1). y = mx + b or 1=-\frac{2}{5} * -3 + b, or solving for b: b = 1-(-\frac{2}{5})(-3).b = -\frac{1}{5}.
  • (2,-1). y = mx + b or -1=-\frac{2}{5} * 2 + b, or solving for b: b = 1-(-\frac{2}{5})(2). b = -\frac{1}{5}.

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

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