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Cloud [144]
2 years ago
10

A parabola with no real zeros, a negative leading coefficient, and an odd y-intercept. pls help!

Mathematics
2 answers:
yuradex [85]2 years ago
6 0

Answer:

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.

pishuonlain [190]2 years ago
5 0

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.

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