Explain why a graph with equation y=2x^2-8x+9 doesn't intersect the x axis?

That's very provable by just determining the discriminant, b^2 - 4ac.

(-8)^2 - 4(2)(9) = 64 - 72 = -8.

You have a negative discriminant, so there aren't any real solutions. That's how you know y will always be positive no matter what x is, and the graph won't ever touch the x-axis, which is the main source for real number solutions.

f ` (x) = 4x - 8 = 0 for turning point

x = 2

f (2) = 8 - 16 + 9 = 1

Turning point (2,1)

f " (x) = 4 > 0

Thus (2,1) is a Minimum turning point

Graph will not cut x axis

One test would be to compute the discriminant. That's the portion of the quadratic formula under the square root symbol.

d = b² - 4ac

If that is negative, then you are adding/subtracting the square root of a negative which has no real solutions. That's equivalent to saying the function has no zeros. Because it has no zeros, it never crosses the line y=0 which is the x-axis.

So let's compute the discriminant:

d = (-8)² - 4(2)(9)

d = 64 - 72

d = -8

As suspected the discriminant is negative and hence the parabola never crosses the x-axis.

Another way to solve this would be to figure out the vertex:

y = 2(x - 2)² + 1

The vertex is (2, 1).

Since the leading coefficient (2) is positive, we have an upward facing parabola whose minimum is the vertex point (2,1). That's above the x-axis so it will never intersect the x-axis.