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Vilka [71]
4 years ago
8

What is the end behavior of the graph of the polynomial function f(x) = –x5 + 9x4 – 18x3?

Mathematics
2 answers:
Anton [14]4 years ago
6 0

<u>Hint-</u>

f(x)=a_0+a_1x+a_2x^2+....+a_nx^n

In the given function,

Leading \ term=a_nx^n \ , \ Leading \ coefficient=a_n \ , \ degree = n

<u>Solution-</u>

The given polynomial,

f(x) = -x^5+9x^4-18x^3

Leading \ term=-x^5 \ , \ Leading \ coefficient=-1 \ , \ degree = 5

The degree of the function is odd and the leading coefficient is negative. So, the end behavior is,

As \ x \rightarrow -\infty,f(x)\rightarrow +\infty \ and \ x \rightarrow +\infty,f(x)\rightarrow -\infty

So the graph will be in 2nd and 4th quadrant.



Helen [10]4 years ago
5 0
Because it's an odd function, the "tails" go off in different directions.  Also, because it's a negative function, the left starts from the upper left and the right goes down into negative infinity.  If it was a positive, the tails would be going in the other directions, meaning that the left would come up from negative infinity and the right would go up into positive infinity.
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The areas of two similar triangles are 98in^2 and 162in^2. What is the ratio of their perimeters when comparing smaller and larg
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\bf \rule{31em}{0.25pt}\\\\ \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{98}}{\sqrt{162}}~~ \begin{cases} 98=2\cdot 7\cdot 7\\ \qquad 2\cdot 7^2\\ 162=2\cdot 9\cdot 9\\ \qquad 2\cdot 9^2 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt{2\cdot 7^2}}{\sqrt{2\cdot 9^2}} \\[2em] \cfrac{s}{s}=\cfrac{7\sqrt{2}}{9\sqrt{2}}\implies \cfrac{s}{s}=\cfrac{7}{9}


bearing in mind that the ratio of the sides, is the same as the ratio of the perimeters.

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3 years ago
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3 years ago
The field inside a running track is made of a rectangle that is 84.39 m long and 73 m wide to gether with a half circle at each
klasskru [66]

Answer:

398.21 m

Step-by-step explanation:

The field inside a running track is made of a rectangle that is 84.39 m long and 73 m wide to gather with a half circle at each end.

So, the radius of each half circle will be \frac{73}{2} = 36.5 m.

Therefore, the total distance around the track i.e perimeter of the area covered by the track will be = 2 × Perimeter of each half circle + 2 × Length of the rectangle.

= Perimeter of full circle + 2 × Length of the rectangle.

= 2\pi r + 2l

= \frac{2 \times 22 \times 36.5}{7} +2 \times 84.39

= 398.21 m.(Answer)

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3 years ago
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