Given the function
![y=\dfrac{3}{5}x+2](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B3%7D%7B5%7Dx%2B2)
and its domain
, its range is ![\{-4,2,5\}](https://tex.z-dn.net/?f=%5C%7B-4%2C2%2C5%5C%7D)
The domain of a function is the set of all possible values that we can input to the function.
The range of a function is the set of all possible values that a function can output.
For the function
![y=\dfrac{3}{5}x+2](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B3%7D%7B5%7Dx%2B2)
if the domain is the set
, we can get the range by substituting each value in the domain into the function. The resulting set of values gives the range of the function.
The range will then be
![\left\{\dfrac{3}{5}(-10)+2, \dfrac{3}{5}(0)+2,\dfrac{3}{5}(5)+2\right\}\\\\=\left\{3(-2)+2, 3(0)+2, 3(1)+2\right\}\\\\=\left\{-4, 2, 5\right\}\\\\](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cdfrac%7B3%7D%7B5%7D%28-10%29%2B2%2C%20%5Cdfrac%7B3%7D%7B5%7D%280%29%2B2%2C%5Cdfrac%7B3%7D%7B5%7D%285%29%2B2%5Cright%5C%7D%5C%5C%5C%5C%3D%5Cleft%5C%7B3%28-2%29%2B2%2C%203%280%29%2B2%2C%203%281%29%2B2%5Cright%5C%7D%5C%5C%5C%5C%3D%5Cleft%5C%7B-4%2C%202%2C%205%5Cright%5C%7D%5C%5C%5C%5C)
Learn more about function <em>domain </em>and <em>range</em><em> </em>here brainly.com/question/13824428
Answer: B
Step-by-step explanation:
Answer:
b i think
Step-by-step explanation:
Given:
The function is
![f(x)=\dfrac{4x}{x^2-16}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B4x%7D%7Bx%5E2-16%7D)
To find:
The asymptotes and zero of the function.
Solution:
We have,
![f(x)=\dfrac{4x}{x^2-16}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B4x%7D%7Bx%5E2-16%7D)
For zeroes, f(x)=0.
![\dfrac{4x}{x^2-16}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B4x%7D%7Bx%5E2-16%7D%3D0)
![4x=0](https://tex.z-dn.net/?f=4x%3D0)
![x=0](https://tex.z-dn.net/?f=x%3D0)
Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.
![x^2-16=0](https://tex.z-dn.net/?f=x%5E2-16%3D0)
![x^2=16](https://tex.z-dn.net/?f=x%5E2%3D16)
Taking square root on both sides, we get
![x=\pm \sqrt{16}](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Csqrt%7B16%7D)
![x=\pm 4](https://tex.z-dn.net/?f=x%3D%5Cpm%204)
So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.
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