h is the distance on the x axis between two points on a curve (a curve in the y=x^2+mx+b format) which form a secant line
we can find the slope of a tangent line of any point on the curve by finding the derivative:
f'(a)=[f(a+h)-f(a)]/h, as h approaches 0

so for y=x^2-8x+9, the derivative will be
= ([(a+h)^2-8(a+h)+9]-[a^2-8a+9])/h
= (a^2+2ah+h^2-8a-8h+9-a^2+8a-9)/h
= (2ah+h^2-8h)/h =(2a+h-8)
= 2a-8

the derivative of the curve y=x^2-8x+9 is 2x-8
This will tell you the slope of the tangent line for any point on that curve by plugging the x value of that point into the derivative. So, for the point (3,-6) on the curve y=x^2-8x+9, you just plug the x value (3) into the derivative (which is 2x-8), which gives you -2.
So the slope of the tangent line at (3,-6) is -2.

You now know derivatives, that's half of calculus.