Question

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Answer 1

it just makes eliminate the need for the general quadratic formula when finding the zeros... it also tells you where the vertex is on the graph. The situation of imaginary roots means the curve is positive definite or negative definite...so it means the grah doesn't cross the x axis.

using your above example... when completing the square you'll have

\[y = (x-2)^2 + 4\]

so the vertex is (2, 4)

and the parabola is concave up...

if you are looking for solutions

set y = 0 and solving for x you have

\[-4 = (x-2)^2\]

then

\[\pm \sqrt{-4} = (x -2) \]

etc...

as the solutions... so a little easier to manipulate

hope this helps

Answer 2

yes... as you get the vertex, and you don't really need intercepts as the curve is always above or below the x axis with imaginary roots