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zhannawk [14.2K]
3 years ago
11

Graph the following piecewise function on a separate piece of graph paper and upload your graph below.

Mathematics
2 answers:
Alecsey [184]3 years ago
7 0
ANSWER

To graph the function

f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x

follow the steps below.

1. Find y- intercept by plugging in x=0.

x=0 is on the interval,  -4\leq x, so we substitute in to

f(x)=x+4

\Rightarrow f(0)=0+4

\Rightarrow f(0)=4

Hence the y-intercept is (0,4)

2. Find x-intercept by setting f(x)=0

This implies that

x+4=0, on -4\leq x

or

2x-1=0 on 3\leq x

We now solve for x on each interval,

x=-4, on -4\leq x

or

x=\frac{1}{2} on 3\leq x

But observe that

x=\frac{1}{2} does not belong to 3\leq x

This means it  can never be an intercept for this piece-wise function.

Hence our x-intercept is (-4,0)

3. Plotting the boundaries of the interval.

For f(x)=x+4 on  -4\leq x

f(-4)=-4+4

\Rightarrow f(-4)=0.

This point (-4,0) coincides with the x-intercept.

f(3)=3+4

f(3)=7

So we have the point (3,7). But note that x=3 does not belong to this interval so we plot this point as a hole.

For f(x)=2x-1 on 3\leq x

f(3)=2(3)-1

\Rightarrow f(3)=5

So we plot (3,5)

f(6)=2(6)-1

\Rightarrow f(6)=11

So we plot (6,11) also as a hole.

Plotting all these points we can now graph the function,

f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x

See attachment for graph.

Contact [7]3 years ago
6 0
For this case we have the following functions:
 For - 4  \leq x  \ \textless \  3 : 
 y = x + 4

 For 3  \leq x \ \textless \  6 :
 y = 2x - 1

 What we need to know for this case is:
 Both functions are one lines
 Both functions have a positive slope
 Both functions are in a certain interval
 Answer:
 See attached image to see functions

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The simplified for of given statement.

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You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of \displaystyle y = -29sin\:\frac{\pi}{6}x + 44\frac{1}{2},in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>cosine</em> graph [photograph on the left], accourding to the <u>horisontal shift formula</u> above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [centre photograph] is shifted \displaystyle 3\:unitsto the right, which means that in order to match the <em>co</em><em>sine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle 3\:units,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{3} = \frac{-\frac{\pi}{2}}{\frac{\pi}{6}}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2}.Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit \displaystyle [12, 15\frac{1}{2}],from there to the y-intercept of \displaystyle [0, 15\frac{1}{2}],they are obviously \displaystyle 12\:unitsapart, telling you that the period of the graph is \displaystyle 12.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 44\frac{1}{2},in which each crest is extended <em>twenty-nine </em><em>units</em> beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- the cosine graph in the photograph farthest to the right is the OPPOCITE of the cosine graph in the photograph farthest to the left, and the reason for this is because of the <em>negative</em> inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of <em>any</em> trigonometric equation, the whole graph reflects over the<em> midline</em>. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

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