i would like the answer

Question

Consider the following polynomial function.

f(x)=(x+3)²(x−1)³(x−2)

Find the zero(s) at which f "flattens out". Express the zero(s) as ordered pair(s).

Answer 1

(-3,0), (1,0), (2,0)

-3 is a zero with multiplicity 2 where the curve "flattens" or has slope=0

for an instant as it's tangent to the x axis at (-3,0), a local minimum

the graph has one other local min a little lower

and one local max much higher where the slope also=0

1 is a zero with multiplicity 3, where the curve intersects the x axis (1,0) is an inflection point

2 is a zero with multiplicity 1 where the curve intersects the x axis at (2,0)

the graph is sort of W shaped, but not so symmetrical

the multiplicity = the exponent of each factor

set each factor =0 and solve for x to find the zero

x+3=0

x =-3

x-1 = 0

x= 1

x-2 =0

x= 2

zeros are -3,-3,1,1,1, and 2

or just -3,1 and 2

multiplicity means the zero repeats

(x+3)^2(x-1)^3)(x-2)

= (x+3)(x+3)(x-1)(x-1)(x-1)(x-2)

set each factor equal to zero and solve for x to get the zeros

f(x) =(x^2+6x+9)(x-2)(x^3+3x^2-3x-1)

= (x^3+4x^2-3x-18)(x^3+3x^2-3x-1)

f'(x)=(x+3)^2(x-1)(3)(x-1) + (x-1)^3[(x+3)+(x-1)2(x+2)]

=

Answer 2

The graph kisses the x-axis at x = -3.

The graph has an inflection point at x = 1

Answer 3

Hi Joana,

"Flattens out" isn't a real math term. if you look at the graph in this link:

https://www.desmos.com/calculator/dq1ubxcjur

You will see that the zero that has multiplicity 3 crosses differently than the zero of multiplicity 1, and that the zero of multiplicity 2 is a touch point. Zeros with odd multiplicity greater than 1 don't cross directly over the x axis the way a line would, but bend as they approach the axis and make a shallow crossing. This could be described as "flattening out."

Enjoy