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Describe the graph of the function at its roots. f(x) = (x − 2)3(x + 6)2(x + 12) At x = 2, the graph crossesdoes not intersecttouches the x–axis. At x = −6, the graph crossesdoes not intersecttouches the x–axis. At x = −12, the graph crossesdoes not intersecttouches the x–axis.

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This type of problem can be decided by looking at whether the function is positive or negative in the vicinity of the zeros. There are three roots: -12, -6, and 2 at which the function is clearly 0. Look at the four intervals the x axis:

interval x<-12: function is negative (just test using a number <-12, say, -13)

interval (-12,-6): function is positive

interval (-6, 2): function is negative

interval x>2: positive

From the above you deduce that at the roots the function must be crossing the x-axis (as opposed to just touching it) because the function value changes its sign every time.

For the graph of given function ,

At x=2, the graph crosses x axis

At x=-6, the graph touches x axis

At x=-12, the graph crosses x axis

Given :

Equation of a function  [tex]f\left(x\right)\:=\:\left(x\:-\:2\right)^3\left(x\:+\:6\right)^2\left(x\:+\:12\right)\:[/tex]

Lets find out the roots and analyze

lets set each factor =0 and solve for x

Exponent in each factor tell us the multiplicity . Using multiplicity we check whether crosses or touches x axis

When multiplicity is odd then graph crosses x axis .

when multiplicity is even then graph touches x axis.

[tex]f\left(x\right)\:=\:\left(x\:-\:2\right)^3\left(x\:+\:6\right)^2\left(x\:+\:12\right)\:\\(x-2)^3= 0\\x-2=0\\x=2[/tex]

Root is x=2  with multiplicity 3. 3 is odd

At x=2, the graph crosses x axis

[tex](x+6)^2=0\\x+6=0\\x=-6[/tex]

At x=-6, multiplicity is 2 that is even .

At x=-6, the graph touches x axis

[tex]x+12=0\\x=-12[/tex]

x=-12  with multiplicity 1. 1 is odd

At x=-12, the graph crosses x axis

Learn more : brainly.com/question/11314797