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Sladkaya [172]
1 year ago
7

Determining end behavior and intercepts to graph a polynomial function.Also for part (a) it asks it it falls to the left rises r

ight or falls right rises left or falls both or rises both.

Mathematics
1 answer:
Natali5045456 [20]1 year ago
7 0

In this problem, we must analyze the behaviour of the function:

f(x)=(x+2)^2\cdot(x-1)^2.

(a) Plotting the function, we get the following graph:

We see that the end behaviour of function is: rises on both sides.

(b) By looking at the expression of the polynomial f(x), we see that it has:

• a zero at x = -2 with multiplicity 2,

,

• a zero at x = 1 with multiplicity 2.

The graph a polynomial has the following behaviour according to its zeros:

Using this data, we conclude that the function:

• do not cross the x-axis,

,

• touches but do not cross the x-axis at x = -2, 1.

(c) The y-intercept is the value of y = f(0), the value of f(x) when x = 0:

f(0)=(0+2)^2\cdot(0-1)^2=4\cdot1=4.

(d) The function has:

• zeros of order two at x = -2 and x = 1, so it touches but does not cross the a-axis there,

,

• y-intercept at y = 4.

The graph of the function and the x-intercept and y-intercept points is:

Answer

(a) rises on both sides

(b)

• do not cross the x-axis

,

• touches but do not cross the x-axis at x = ,-2, 1

(c) y-intercept = 4

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Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

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let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

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so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

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