What is the equation of the parabola that has a vertex at # (4, 6) # and passes through point # (-7,-8) #?

Known: The equation standard form is #y=ax^2+bx+c#

Using the two points to give simultaneous equations

#-8=a(-7)^2+b(-7)+c#.................................(1)
#" "6=a(4)^2+b(4)+c#............................................(2)

Known that #x_("vertex")=(-1/2)(b/a)#

#=>4=(-1/2)(b/a)#

So #a(2xx4)=-b#

#=> b=-8a#....................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Substitute (3) into (1) and (2)")#

#-8=a(-7)^2+(-8a)(-7)+c#
#" "6=a(4)^2+(-8a)(4)+c#

#-8=49a+56a+c#
#" "6=16a-32a+c#

#-8=" "105a+c" "...................(1_a)#
#" "6=-16a+c" "................(2_a)#

#color(brown)("Subtract "(2_a)" from "(1_a))#

#-14=121a#

#color(blue)(a=-14/121)#
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(=> b=-8a -> b= (-8)xx(-14/121) =+112/121)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Substitute for "a" in "(2_a))#

#" "6=-16(-14/121)+c#

#color(blue)(c=4 18/121 =502/121)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The equation of a parabola with vertical axis can be expressed in vertex form as:

#y = a(x - h)^2 + k#

where #(h, k)# is the vertex and #a# a multiplier.

In our example, the equation of the parabola can be written in the form:

#y = a(x - 4)^2 + 6#

from which we can deduce:

#a = (y-6)/(x-4)^2#

Since we want our parabola to pass through #(-7, -8)# substitute these values of #x# and #y# into this equation to find:

#a = (-8-6)/(-7-4)^2 = -14/121#

So the equation of our parabola may be written:

#y = -14/121 (x-4)^2 + 6#

#=-14/121 x^2 + 112/121 x - 224/121 + 6#

#=-14/121 x^2 + 112/121 x +502/121#

graph{-14/121 x^2 + 112/121 x +502/121 [-20, 20, -10, 10]}