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amid [387]
3 years ago
6

Marlon and Michelle have to graph a function with the equation y=x2and are having to debate about which is the most important el

ement to get started, the domain or the range . which of them is correct?

Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

Correct answer is the domain.

Step-by-step explanation:

A domain is the input set of a function.

A range is the output set of the function.

It is more  important to know the domain of the function to be able to determine the corresponding output set so the function can be graphed. To start with the domain is very useful because every domain element has a corresponding unique output element.

Suppose you started with the range element of 9. The input set for an output of 9 is {-3,3}. This makes it hard to match up the elements. This example highlights why it is important to start with the domain rather than the range.

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I hope this helps you



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3 years ago
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The rabbit population in a certain area is 200% of last year's population.There are 1,000 rabbits this year.how many were there
alexandr1967 [171]

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Find the volume of the rectangular prism.<br> The volume is cm?<br><br> 8 cm<br> 5 cm<br> 9 cm
dedylja [7]

Answer:

360 cm

Step-by-step explanation:

volume= length x width x height

8*5*9=360 cm

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Step-by-step explanation:

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3 years ago
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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20
vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of y=f(x) is a vertical has an asymptote at x=a if at least one of the following statements is true:

1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty

The function is:

f(x)=\frac{x^2+7x+10}{x^2+9x+20}

First of all, let't factor out:

f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq  5

From here:

\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty

\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty

Accordingly:

x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

5 0
4 years ago
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