Thread: Polynomial 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...

Originally Posted by missionmom

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...

Because a < 0 in your quadratic equation this graph will open upwards.
Therefore f(x) > 0 between the two roots and hence f(x) < 0 when x < -1.52 and x > 3.87

doesnt it open down because it has a negative"a" value but a positive exponent?

Originally Posted by missionmom

doesnt it open down because it has a negative"a" value but a positive exponent?

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