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Ulleksa [173]
2 years ago
8

Solve the word problem using the plotted points on the graph.

Mathematics
1 answer:
LuckyWell [14K]2 years ago
7 0

Answer:

In the graph, we can see that the relation between length and weight is given by the adjusted line, which passes through the points (24, 16) and (28, 25)

Remember that a linear relation can be written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

If this line passes through the points (x₁, y₁) and (x₂, y₂) the slope can be calculated as:

a = (y₂ - y₁)/(x₂ - x₁)

Then in our case, the slope will be:

a = (25 - 16)/(28 - 24) = 9/4

y = (9/4)*x + b

Knowing that this line passes through (24, 16), we know that when x = 24, y must be equal to 16.

If we replace these in the equation, we can find the value of b.

16 = (9/4)*24 + b

16 = 54 + b

16 - 54 = b - 38

Then the equation is:

y = (9/4)*x - 38

Now that we know the equation, we can simply replace y by 34 pounds to find the value of x.

34 = (9/4)*x - 38

34 + 38 = (9/4)*x

72*(4/9) = x = 32

So we can estimate that the length of a fish that weighs 34 pounds is 32 (I do not know the unit of length, I can't see the horizontal axis on the image)

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b = 9

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Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but x \ne 0</h3>

To write this in interval notation, we can say (-\infty, 0) \cup (0, \infty) which is the result of poking a hole at 0 on the real number line.

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The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say (-\infty, \infty) which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

=======================================================

Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation x = \pm 4

These are the locations where the blue hyperbolic curve crosses the x axis.

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Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

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Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

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The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer:  Oblique asymptote is y = 3x</h3>
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