Hi:) is it possible to plot a graph with non-linear and linear equations?
2 answers:
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9514 1404 393
Answer:
possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}
actual roots: -1, (2 ±4i√2)/3
no turning points; no local extrema
end behavior is same-sign as x-value end-behavior
Step-by-step explanation:
The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.
The Rational Root Theorem tells us any rational roots will be of the form ...
±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}
= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots
Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.
The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.
The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):
f(x) = (x +1)(3x^2 -4x +12)
The complex roots of the quadratic can be found using the quadratic formula:
x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6
x = (2 ± 4i√2)/3 . . . . complex roots
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The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.
