Question-Graph the polynomialf(x)=-x^4+3x^3+4x^2-3x+2 using a graphing calculator. Find the interval wheref(x)<=0?

So found that the real zeros were x=-1.52 and 3.87

I dont understand how to find where f(x)<=0 from there.

Please explain...

Printable View

- September 1st 2009, 12:29 PMmissionmomPolynomial equation and graphing calculator?
Question-Graph the polynomial

*f*(x)=-x^4+3x^3+4x^2-3x+2 using a graphing calculator. Find the interval where*f*(x)<=0?

So found that the real zeros were x=-1.52 and 3.87

I dont understand how to find where f(x)<=0 from there.

Please explain... - September 1st 2009, 12:43 PMe^(i*pi)
- September 1st 2009, 12:50 PMmissionmom
doesnt it open down because it has a negative"a" value but a positive exponent?

- September 1st 2009, 12:57 PMe^(i*pi)
No it opens up, it's convention to treat it a -a(x^2), I think it's something to do with order of operations too

This isn't a quadratic though >.<, the above point still applies though

I think the best thing to do is to plug 3 values one each side of the root and in the middle, for example f(-2), f(0) and f(4). The results of these should help you pick