Question 200749
The graph of this function {{{f(x) = 8x^2-2^4}}}will give you the answers:
{{{graph(400,400,-5,5,-5,8,8x^2-2x^4)}}}
But you can also get the answers algebraically:
{{{f(x) = 8x^2-2x^4}}} Substitute {{{y = f(x)}}}
The roots (zeros) occur at y = 0, so substitute this...
{{{8x^2-2x^4 = 0}}} Solve for the x's, there should be four roots for this 4th degree polynomial. Factor out {{{2x^2}}}
{{{(2x^2)(4-x^2) = 0}}} Apply the zero product rule so that...
{{{2x^2 = 0}}} or {{{4-x^2 = 0}}} 
For {{{2x^2 = 0}}}, {{{x = 0}}} There are two roots at x = 0.
For {{{4-x^2 = 0}}}, {{{x^2 = 4}}} then {{{x = 2}}} and {{{x = -2}}}
So the four roots (zeros) are:
{{{highlight(x = 0)}}}
{{{highlight_green(x = 0)}}}
{{{highlight(x = 2)}}}
{{{highlight_green(x = -2)}}}
Compare these with the graph!
The inflection points (where the curve changes sign) are:
{{{x = -sqrt(2)}}}
{{{x = 0}}}
{{{x = sqrt(2)}}}
My appologies for the first answer about the points of inflection!