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

2 Answers
Jun 22, 2017

Answer:

See below

Explanation:

#g(x) = x^3-3x^2#

Firstly we are asked to find the zeros of #g(x)#

#g(x) = 0 -> x^3-3x^2=0#

#x^2(x-3)= 0#

#:. x =0 or x= 3# are the zeros of #g(x)#

To graph #g(x)# it now becomes necessary to find the turning point(s). Since this question is in "Precalculus" section we will observe the graph of #g(x)# (below) rather than analysing the function using calculus.

graph{x^3-3x^2 [-10, 10, -5, 5]}

From the graph, we can see the zeros at #x=0# and #x=3# and the local minimum at #(2, -4)#

Jun 22, 2017

Answer:

#"see explanation"#

Explanation:

#" set " g(x)=0" for zeros"#

#x^3-3x^2=0rArrx^2(x-3)=0#

#x=0" multiplicity 2"#

#x=3" multiplicity 1"#

#"we require more than the zeros to draw the graph"#

#"differentiate " g(x)" for stationary points and nature"#

#g'(x)=3x^2-6x#

#"equate to zero for stationary points"#

#3x^2-6x=0rArr3x(x-2)=0#

#"stationary points at " x=0" and " x=2#

#g(0)=0rArr(0.0)" is a stationary point"#

#g(2)=8-12=-4rArr(2,-4)" is a stationary point"#

#"using "color(red)"second derivative test"#

#g''(x)=6x-6#

#g''(0)=-6rArr(0,0)color(red)" is a maximum"#

#g''(2)=6rArr(2,-4)color(red)" is a minimum"#

#"since degree of " g(x)" is odd and the leading "#
#"coefficient is positive"#

#"then graph starts down and ends up"#

#"combining the above gives the graph of " g(x)#
graph{x^3-3x^2 [-10, 10, -5, 5]}