A standard question in Sec 3, the idea goes like this.

If a curve is always negative for all values of x, it means that no matter our choice of values of x, the y-value of the graph is negative, so that the graph will be entirely below the x-axis. It’s something like sit and reach in NAPFA test. No matter how you try to push all the way to the other side, you can’t reach it (unless you are a sit and reach master).

In such situations, the graph will never ever cut the x-axis, leading to no point of intersection between the graph and the x-axis. As such, this is taken to be “no real roots”, and b2 - 4ac < 0.

Only sad face graphs can exhibit the properties of “entirely below the x-axis”. Because howsoever you draw a “smiley” face graph, there will always be portions of the graph above the x-axis.

Sad face graphs are denoted by a negative coefficient of x2, such as -1.

As such, -a must be negative, and therefore a must be positive.

Sorry, never realised that in the corrections you shifted everything to make ax2 + ... > 0.

Again, the fact that ax2 + ... > 0 for all values of x means that the graph will always be above (and never intersect) the x-axis, leading to no real roots and as such b2 - 4ac < 0.

The difference this time is that the coefficient of x2 must be positive, since only “smiley” face graphs are capable of being entirely above the x-axis.

As such, a must be positive.

The condition that the coefficient of x2 must be positive or negative is often missed out by students, so do take note.

One more thing. I need you to distinguish between the original graph and the discriminant (b2 - 4ac) graph. The original graph will be drawn such that the graph does not touch the x-axis. The b2 - 4ac graph, however, will cut the horizontal axis twice most of the time if it turns out to be quadratic. The one you wrote in blue ink is considered to be the discriminant graph and not the original equation graph.

Thank you soso much!