# Discriminant when graph lies above or below the x axis.

Suppose a quadratic equation has been given where the a value (ax^2 + bx + c) is a positive and it has been said that the graph of the equation lies above the x-axis- what is the discriminant?

For example- 2x^2 + kx - 5 = y; the graph lies above the x-axis- find the possible values of k. I know how to solve it, just not sure what discriminant to take.

In the same sense, if the a value is a negative, and the graph is said to be lying above the x axis, is the discriminant > 0 ?

You need the discrimination to be negative so that the quadratic function has no real roots.

$$\Delta < 0 \iff b^2-4ac < 0$$

Therefore, the graph either lies entirely above or below the $$x$$-axis. This is determined by the value of $$a$$.

If $$a > 0$$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $$x$$-axis.

If $$a < 0$$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $$x$$-axis.

For your example, you have $$y = 2x^2+kx-5$$, so you use

$$k^2-4(2)(-5) < 0 \iff k^2+40 < 0 \iff k^2 < -40$$

Clearly, the inequality is not true for real values of $$k$$, since $$k^2 \geq 0$$, so no value of $$k$$ allows the entire graph to remain above the $$x$$-axis.

• but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0? – De Sith Dec 9 '18 at 14:41
• A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $\Delta < 0$.) – KM101 Dec 9 '18 at 14:45
• yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values? – De Sith Dec 9 '18 at 14:50
• Yes, the discriminant would be positive then, since there are two real solutions. – KM101 Dec 9 '18 at 14:52

If the graph of a quadratic function of $$x$$ lies above or below the $$x$$-axis, in order words, fails to have real roots, then the discriminant must be negative, since in this case the square root would be imaginary.

Thus, in such a case for your function $$ax^2+bx+c,$$ it is the case that $$b^2-4ac<0.$$