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IgorC [24]
3 years ago
5

the change in water vapor in a cloud is modeled by a polynomial function, c (x). describe how to find the x - intercepts of c (x

) and how to construct a rough graph of c (x) so that that the meteorologist can predict when there will be no change in water vapor. you may create a sample polynomial to be used in your explanation.

Mathematics
2 answers:
lukranit [14]3 years ago
8 0

The change in the water vapors is modeled by the polynomial function c(x). In order to find the x-intercepts of a polynomial we set it equal to zero and solve for the values of x. The resulting values of x are the x-intercepts of the polynomial.

Once we have the x-intercepts we know the points where the graph crosses the x-axes. From the degree of the polynomial we can visualize the end behavior of the graph and using the values of maxima and minima a rough sketch can be plotted.

Let the polynomial function be c(x) = x ² -7x + 10

To find the x-intercepts we set the polynomial equal to zero and solve for x as shown below:

x ² -7x + 10 = 0

Factorizing the middle term, we get:

x ² - 2x - 5x + 10 = 0

x(x - 2) - 5(x - 2) =0

(x - 2)(x - 5)=0

x - 2 = 0 ⇒ x=2

x - 5 = 0 ⇒ x=5

Thus the x-intercept of our polynomial are 2 and 5. Since the polynomial is of degree 2 and has positive leading coefficient, its shape will be a parabola opening in upward direction. The graph will have a minimum point but no maximum if the domain is not specified. The minimum points occurs at the midpoint of the two x-intercepts. So the minimum point will occur at x=3.5. Using x=3.5 the value of the minimum point can be found. Using all this data a rough sketch of the polynomial can be constructed. The figure attached below shows the graph of our polynomial.

Levart [38]3 years ago
3 0
Let the polynomial be:
c(x) = x² - 5x + 6
The x-intercepts, 2 and 3, represent the points where there will be no water vapor in the clouds. Using the x intercepts and the vertex of the equation, we can plot a graph to see how the water vapor changes.
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Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


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k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

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→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

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