Answer:
Step-by-step explanation:
A parabolic graph with no real zeros never touches or crosses the x-axis.
If the leading coefficient is negative, that tells us that the parabola opens down (and, in this case, that the entire graph is below the x-axis, because there are no real zeros.
If the y-intercept is odd, then the constant term, c, in ax^2 + bx + c, must be a negative, odd integer.
Suppose we say that the vertex of a particular parabola that is described by the above is (-2, -1) and that the y-intercept is -3. (Note: this is strictly an example.) Then the vertex equation of this parabola is obtained from the general vertex equation of a parabola,
y = a(x - h)^2 + k. Here h = -2 and k = -1, and so the particular parabola is
y = -a(x + 2)^2 - 1, where a is a constant and -a is negative.
In general form, this would be y = -a(x^2 + 4x + 4) - 1, or
y = -ax^2 - 4ax - 4a -1
and the constant terem (-4a -1) must be an odd negative integer
This is only one of an infinite number of possible equations for the parabola described above.
