Hi my good friends!,

I am going through a paper with memo, and came upon a problem which I just cannot grasp. Please would someone explain this to me?

We have a parabola drawn above the x- axis with the axis of symetry to the right of the y-axis. The left side intersects Y at p.

The function is: \(f(x)=2x^2-3x+p\)

The question is: Detrmine the value of p for which the graph will always be above the x- axis.

The answer is this, but I need someone to explain this to me please:

\(\Delta = b^2-4ac\)

\(=(-3)^2-4(2)(p)\)

\(=9-8p\)

I understand the discriminant part....but why is this now put smaller than 0?

\(9-8p<0\)

\(P> {9 \over 8}\)

How would I teach / explain this to a pupil?.Thank you all very much for taking the time to explain this to me...Thank you.

juriemagic Sep 28, 2019

#1**+4 **

If the discriminant is < 0, then we will have non-real roots. In other words, the graph of f(x) will never intersect the x axis. [ It will always be above the x axis ]

This might help : https://www.desmos.com/calculator/nlbuvyisgg

When p = 8/8 (or, 1) the graph intersects the x axis twice

When p = 9/8 the graph is tangent to the x axis

When x = 10/8 the graph is above the x axis

CPhill Sep 28, 2019

#3**+1 **

I had a look at the link you provided, thank you kindly...I see it now...have a blessed day..

juriemagic
Sep 28, 2019