+0

# Value of p

0
88
3
+597

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.

Sep 28, 2019

#1
+104962
+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

Sep 28, 2019
#2
+597
+1

CPhill,

Thank you for this. I do appreciate...

juriemagic  Sep 28, 2019
#3
+597
+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