If a quadratic equation has solutions #x=a# and #x=b#, then #x-a=0# and #x-b=0#. Furthermore, the quadratic can be written as #y=c(x-a)(x-b)#, where #c# is some constant. The reasoning is that if you set #y# equal to #0#, you get:
#c(x-a)(x-b)=0#
Which is the same as:
#(x-a)(x-b)=0#
And so the solutions are #x=a# and #x=b# - which is exactly what we started with.
Alright, enough theory - let's get on with it! We are told that the #x#-intercepts are #-3# and #2#, and since #x#-intercepts are the same thing as zeros, #x=-3# and #x=2# are solutions. Following the process from above, we can write the quadratic as:
#y=c(x+3)(x-2)#
To solve for #c#, we use the other piece of info we were given: the point #(3,6)#:
#y=c(x+3)(x-2)#
#->6=c(3+3)(3-2)#
#->6=c(6)(1)#
#->6=6c->c=1#
So the equation of the quadratic is:
#y=1(x+3)(x-2)#
#->y=(x+3)(x-2)=x^2+3x-2x-6=x^2+x-6#
