What are the ordered pairs of thesolutions for this system of equations? f(x) = x^2 – 2x + 3; f(x) = -6x
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Answer:
x < -3 or x > 4
Step-by-step explanation:
The product of the factors is 4th-degree, and the leading coefficient is 1 (positive).
The factors tell you the following about the zeros:
x = -3 . . . sign change from + to -
x = -1 . . . graph touches, but no sign change. - on either side of x=-1
x = 4 . . . sign change from - to +
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The function will be positive for x < -3 and for x > +4.
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<em>Additional comments</em>
The value of the function is zero when any of its factors is zero. The value of a factor changes sign when the value of x changes from one side of the 0 to the other. For example, here, we have x+1 as a factor, so x=-1 is a zero. When x = -0.9, the factor is positive (-0.9 +1 = 0.1). When x = -1.1, the factor is negative (-1.1 +1 = -0.1). If this factor were to the first power, it would cause the value of the function to change sign at x=-1.
However, the factor (x+1) has an even power: (x+1)^2. That means when the factor is negative, its square is positive, and when the factor is positive, its square is positive. In short, the factor (x+1)^2 causes the function to be zero at x=-1, but does not make the function change sign there.
The product of all of these factors will result in a polynomial with x^4 as the highest-degree term. That means the function is of even degree. The leading coefficient is 1, so is positive, and the function will generally have a U-shape. The left-most (odd-degree) zero will be where the function changes sign from positive to negative. The right-most (odd-degree) zero will be where the function changes sign from negative to positive. Those zeros are x=-3 and x=+4, respectively. There are no places between these where the function changes sign, so these zeros are the ends of the regions where the function is positive (>0).
