In an equation #y=a(x-h)^2+k#, we have vertex as #(h,k)# and intercepts on #y#-axis can be found by putting #x=0# and intercepts on #x#-axis can be found by putting #y=0#. Let us convert the given equation in vertex form.

#12y=x^2-6x+45# can be written as

#y=1/12(x^2-6x)+45/12#

or #y=1/12(x^2-2xx3xx x+3^2-3^2)+45/12#

or #y=1/12(x^2-2xx3xx x+3^2)-9/12+45/12#

or #y=1/12(x-3)^2+36/12#

or #y=1/12(x-3)^2+3#

Hence vertex is #(3,3)#

intercept on #y#-axis is #y=15/4=3 3/4#

Observe that as #a# and #k# are positive, #y# is always greater than #0# and can never be #0#, hence no #x#-intercept.

graph{12y=x^2-6x+45 [-7.12, 12.88, -0.4, 9.6]}