How do you convert from vertex form to intercept form of ##y-4=-(x-4)^2##?

The representation of the original function in intercept form is ##y=-(x-2)(x-6)##

Intercept form of a quadratic function, by definition, is a form ##y=k(x-alpha)(x-beta)## It's called intercept form because ##alpha## and ##beta## are values of ##x## where ##y## equals to zero and, therefore, values where parabola that represents a graph of this quadratic function intercepts the X-axis.

In other words, ##alpha## and ##beta## are solutions to an equation ##k(x-alpha)(x-beta)=0##

Transform our expression into traditional functional form. ##y-4=-(x-4)^2## ##y=-x^2+8x-12## Now let's find the solutions of the equation ##-x^2+8x-12=0## or, in a simpler representation, ##x^2-8x+12=0## Solutions are ##x_1=2, x_2=6##.

Therefore, representation of the original function in intercept form is ##y=-(x-2)(x-6)##

The graph of this function follows (notice the points where it intercepts the X-axis are ##x=2## and ##x=6##). graph{ -(x-4)^2+4 [-10, 10, -5, 5]}

The original form of this function ##y-4=-(x-4)^2## is called vertex form because it tells the location of the vertex of the parabola that represents a graph of this quadratic function - point ##(4,4)##.

It can be easily seen if it is written as ##y=-(x-4)^2+4##. In this case, the rules of graph transformation tell us that the prototype function ##y=-x^2## with vertex at point ##(0,0)## and "horns" directed down should be shifted by ##4## to the right to represent function ##y=-(x-4)^2## and then by ##4## up to represent function ##y=-(x-4)^2+4##. These two transformations shift the vertex to point ##(4,4)##.