+0  
 
0
187
3
avatar+603 

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
avatar+109450 
+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

 

 

 

cool cool cool

 Sep 28, 2019
 #2
avatar+603 
+1

CPhill,

 

Thank you for this. I do appreciate...

juriemagic  Sep 28, 2019
 #3
avatar+603 
+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

20 Online Users