given the polynomial function of P(x)=3x^3+2x^2-4
2007-10-12 04:13:06 · 3 answers · asked by aprille cute 1 in Science & Mathematics ➔ Mathematics
You simply substitute 0 for the x or y, depending on which zeros you want to find. so in the above equation.
substituting 0 for y you get : 0=3x^3+2x^2-4
and substituting 0 for x: P(x)=3(0)^3+2(0)^2-4
If you think about what the zeros really are it's not so hard. The zeros are where the graph of the function cross the x and y axis in the cartesian plane. So what is your x value when y is at zero and what is your y value when x is at zero. Keeping in mind there can be more than one zero intercept, which in this case there are.
2007-10-12 04:29:55 · answer #1 · answered by Martin S 2 · 0⤊ 1⤋
P(x)=3x^3+2x^2-4
This has a real root at approximmately .917
The other two roots are imaginary.
2007-10-12 04:30:40 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
x=0.917 is close to 0.
The root lies between 0.917 and 0.918.
Done by a computer program and trial and error.
2007-10-12 04:42:35 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋