How do I plot the graph of f(x) = 4/(x+3)?

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To plot the graph of f(x) = 4/(x+3), start by determining the function's domain and identifying key limits. Understanding the transformation from y = 1/x to y = 1/(x+3) is essential, as the latter represents a horizontal shift left by 3 units. The coefficient 4 indicates a vertical stretch of the graph. Familiarity with the basic shape of y = 1/x will aid in visualizing the transformed function. This foundational knowledge is crucial for accurately drawing the graph.
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Problem drawing graph, please help!

Hi everyone, I'm struggling to draw the graph of f(x) = 4/(x+3)

I'm just wondering how you go about plotting a graph like this and the process you use to do it. Thanks for your time
 
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Well, first think about the domain of the function. Then think about some limits of interest.

Btw, http://home.alltel.net/okrebs/page192.html" . :smile:
 
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Learn to draw y = 1/x, first then realize what transformation maps y = 1/x on to y = 1/(x+3). The 4 only represents a one-way stretch in the y-direction, doesn't it.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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