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

1 Answer
Apr 2, 2015

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)#.