f(x)=x^2-5x-14 can be written f(x)=x^2-5x+25/4-25/4-56/4=(x-5/2)^2-81/4. This "completes the square". How did I do this? I looked at x^2-5x and asked myself: what do I need to add to this to complete the square involving x? What you do is halve the x term, 5, and then square it: (5/2)^2=25/4. Now I have the complete square of x-5/2. But I have to compensate for adding in 25/4 so I do this by subtracting -25/4, but keep this separate from the perfect square and absorb it into the number -14. I can represent -14 as an improper fraction -56/4 in a convenient form to combine with -25/4 so that -25/4-56/4=-81/4 is the combined result. Now f(x)=(x-5/2)^2-(9/2)^2, and some of the properties of the parabola can be more easily seen. The graph is U-shaped and cuts the x axis at the zeroes for the function. To find these intercepts, we put (x-5/2)^2-(9/2)^2=0 so (x-5/2)^2=(9/2)^2 and taking square roots of each side we get: x-5/2=+9/2, from which x=5/2+9/2 and x=7 or -2 (also f(x)=(x-7)(x+2)). These are x intercepts and halfway between them at x=5/2 (1/2 of 7+(-2)) and y=-81/4 is the vertex (in this case the lowest point with x=5/2 as the vertical axis of symmetry). This makes it easy to visualise and draw the graph. We can also work out the y intercept, when x=0. This is -14. All the points labelled fix the symmetrical parabola into position. The arms of the U spread apart and you can plot various values away from the zeroes to see what the spread is.