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Let P(x) = (x - 3)(x+4)^2(x-1)^3. (a) Find the zeros of P(x) and the multiplicity of each. (b) Determine whether the graph crosses or touches the x-axis at each x-intercept. (c) Determine the end behavior of P(x). (d) Construct the sign chart (e) Sketch the graph of the function.

Solution

a)
P(x) = (x - 3)(x + 4)^2(x - 1)^3

(x - 3)(x + 4)^2(x - 1)^3 = 0

x -3 = 0 , x + 4 = 0 , x - 1 = 0

x =3 , -4 and 1

So, we have :
x = 3, multiplicity = 1
x = -4, mult = 2
x = 1, mult = 3

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b)
At x = 3, it crosses because mult = odd
At x = -4, it touches because mult = even
At x = 1, it crosses because mult = odd

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c)
This is clearly degree 6 and positive leading term...

So, end behavior for even degree and positive leading term is :

Up on both sides

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d)
Zeros were 3 , -4 and 1

Region 1 : (-inf , -4)
Test = -5
With this, (x-3) is negative
(x-1)^3 is also negative
neg * neg = positive...

Region 2 : (-4 , 1)
Test = 0
With this, we get positive again

Region 3: (1 , 3)
Test = 2
With this, we get negative

Region 4 : (3 , inf)
Test = 4
With this, posiitve

So, positive over (-inf , -4) U (-4 , 1) U (3 , inf)
Negative over (1 , 3)