The “h = 3” for part i does not apply for part ii.

We need to start part ii “afresh”.

y = 2x^2 + (h + 4)x + 2h

Suppose y = 0.

2x^2 + (h + 4)x + 2h = 0

Our discriminant of roots for this equation
= (h + 4)^2 - 4 (2) (2h)
= h^2 + 4h + 4h + 16 - 16h
= h^2 - 8h + 16
= (h - 4) (h - 4)
= (h - 4)^2

Wait for my next post.

Our discriminant = (h - 4)^2

As (h - 4)^2 is a squared expression, its lowest value is 0 (occurs when h = 4); for other values of h, (h - 4)^2 will spew out a positive value.

Note that squaring a negative number gets us a positive number, so it’s not possible for the square of a real number to be negative.

In other words,
- our discriminant = 0 when h = 4
- our discriminant > 0 for all other values of h

So, we can have two cases:

- “Two real and equal roots” when h = 4
- “Two real and distinct roots” for all other values of h

When an equation has two real and distinct roots, the corresponding curve will be two intercepts; when this happens, a graph will “cross over” from the positive side to the negative side (for smiley face curves) or the negative side to the positive side (for sad face curves).

Either way, the curve will exist on BOTH the positive and negative portions.

The question wants us to find those h for which the curve cannot be negative. The only way is when the curve touches the x-axis (discriminant = 0) or does not even touch it (discriminant < 0).

As we know it too well…

h = 4 is the only possible way.

In our case, we just need to check that the coefficient of x^2 is positive.

Because only smiley face curves are capable of being “never negative”, we must ensure this.

Our coefficient is thankfully 2, so we have our answer, h = 4.