The discriminant of a polynomial (in this case a quadratic polynomial; degree 2) gives information about the types of zeros the polynomial has.
In the case of a quadratic, `f(x)=ax^2+bx+c` , the discriminant is `b^2-4ac` . Now if:
`b^2-4ac>0` there are two real zeros, two real roots, or two x-intercepts of the graph.
`b^2-4ac=0` there is one real root, one real zero, one x-intercept. This is called a repeated root, or a root of multiplicity 2.
`b^2-4ac<0` there are no real roots, no real zeros, and the graph has no x-intercepts. There are two imaginary roots.
In the problem given, since the graph lies entirely above the x-axis, there are no x-intercepts. Thus the discriminant must be negative, so the answer will be 1)-6.
** One way to remember what the discriminant tells you is to examine the solutions given by the quadratic formula: if `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)` . Note that the line of symmetry for the parabola is `x=(-b)/(2a)` . If the discriminant is zero, then its square root is zero and the solution is the vertex of the parabola. If the discriminant is positive, there are two solutions located the same distance on either side of the axis of symmetry. And if the discriminant is negative, there are no real solutions so the graph cannot cross the x-axis.
If b^2 - 4ac < 0 then the function has no real roots, but I would guess that it might lie either above or below the x-axis.
And the proper answer would be 1) -6.