Question

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

Answer 1

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

Explanation:

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)`.