Graphing polynomials ?Sketch a graph of the following polynomials. identify local extrema, inflection points, and x?- ?and y-intercepts when they exist.

Solution: Step1 Given that f(x)=x^4-6x^2 Differentiate the given equation to find f’(x) we get, f’(x)= 4x^3-6*2x = 4x^3-12x Again differentiate f’(x) to find f’’(x) we get, f’’(x)=4*3x^2-12 = 12x^2-12 Step2 To get extreme values we have to use f’(x)=0 => 4x^3-12x=0 =>4x(x^2-3)=0 => 4x=0, (x^2-3)=0 => x=0, 3,- 3 Critical points are 0,- 3 and + 3 Step3 To find inflection points we have to use f’’(x)=0 =>12x^2-12=0 =>12x^2=12 =>x^2=12/12=1 => x=±1 Evaluate f’’(x) at the critical points we get, f”(0 )=12x^2-12= 12*(0)^2-12 =0-12= -12<0 f(x) has a local maximum at x=0 And f”( 3)= 12x^2-12 = 12*( 3)^2-12 = 12*3-12=36-12=24>0 Has a local minimum at x= 3 And f”( 3)= 12x^2-12 = 12*(- 3)^2-12 = 12*3-12=36-12=24>0 Has a local minimum at x=- 3 Step4 The corresponding function values are f(0 )= x^4-6x^2 = (0)^4-6*(0)^2= 0-0=0 , f( 3 )= x^4-6x^2 =( 3)^4-6*( 3)^2 =9-18=-9 And f(- 3 )= x^4-6x^2 =( - 3)^4-6*( 3)^2 =9-18=-9 Finally we see that f”(x) changes sign at x=-1,1 f(1)=5 and f(-1)=-5 Therefore the inflection point is (1,-5) and (-1,-5) And the local maximum at (0,0) , Local minimum at (- 3,-9) and ( 3,-9) For x-intercepts, we have to solve the equation f(x)=0 =>x^4-6x^2=0 =>x^2(x^2-6)=0 =>x=± 6 Therefore the intercepts are (- 6,0) and ( 6,0) Step5