If P(x) = x^2-4x, then since
P(x) = x(x-4), it has two x-intercepts.
Q(x) = x(x^2-4) = x(x-2)(x+2)
so Q has 3 roots.
x^2-ax = x(x^2-a) = x(x-√a)(x+√a)
so there are 3 roots
With 3 roots, there must be 2 extrema.
x^2+ax = x(x^2+a)
since a>0, x^2+a can never be zero, so there is only 1 x-intercept, at x=0.
There are no extrema, since the derivative is never zero.
How many x-intercepts and how many local extrema does the polynomial P(x)= x(sqrt)3-4x have?
How many x-intercepts and how many local extrema does the polynomial Q(x)=x^3-4x have?
and If a>0, how many x-intercepts and how many local extrema does each of the polynomials P(x)=x^3-ax and Q(x)=x^3+ax? Explain your answer.
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