How do you graph by using the zeros for #g(x)=1/10(x+1)^2(x-3)^3#?

1 Answer
Aug 12, 2018

Answer:

x intercepts : # x=-1, x=3#, y intercept: #y=-2.7#
local minimum at #(0.6,-3.54)#

Explanation:

#g(x)= 1/10(x+1)^2( x-3)^3#

Vertical asymptote occur when denominator is zero.

So no vertical assymptote. Degree is #x^5# , odd and

leading coefficient is #(+)# , so end behavior is "down /up"

i.e # y -> -oo # as #x -> -oo# and

# y -> oo # as #x -> oo#

Zeros are #x= -1# with multiplicity of #2# and

#x= 3# with multiplicity of #3:.# x intercepts are

#x=-1 and x=3#, y intercept: Putting #x=0# in the equation

we get,#y= 1/10*(1)^2* (-3)^3 = -2.7#

#g(x)= 1/10(x+1)^2( x-3)^3#

#g^'(x)= 1/10[2(x+1)(x-3)^3+(x+1)^2*3(x-3)^2]# or

#g^'(x)= 1/10[(x+1)(x-3)^2(2(x-3)+3(x+1)]# or

#g^'(x)= 1/10[(x+1)(x-3)^2(5 x-3)]#

#g'(x)=0 :. # slope is zero at #x=-1,x=3 and x=3/5=0.6#

#g(0.6)~~-3.5389~~ -3.54 # , critical point #x=0.6#

Slope at # x<-0.6 , g^'(x) <0 # i.e decreasing and

slope at # x> -0.6 , g^'(x) >0 # i.e increasing.

So #(0.6, -3.54)# is local minimum.

At #x=-1# the graph will not cross #x# axis as the multiplicity

is even #2# and at #x=3# the graph will cross #x#axis as the

multiplicity is odd #2#.

graph{1/10(x+1)^2(x-3)^3 [-10, 10, -5, 5]}[Ans]