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dezoksy [38]
2 years ago
10

Find the domain and range of the following graph.

Mathematics
1 answer:
RideAnS [48]2 years ago
4 0

Domain:

From left-to-right, the graph starts at x = -4 and goes on forever to the right.

The domain is x ≥ -4.

Range:

From bottom-to-top, the graph gets as low as y = 1 and goes up forever.

The range is y ≥ 1.

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Rationalize the denominator.
Maslowich
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166412

——————————
 
     \mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}


Multiply and divide by the conjugate of the denominator, which is \mathsf{\sqrt{a}+2\sqrt{y}:}

     \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:
 
     \mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}} 

     \mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

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