What is the vertex of #y=5(x+3)^2-9 #?

There are two ways to solve it:

1) Quadratics:

For the equation #ax^2+bx+c=y# :

The #x#-value of the vertex #=(-b)/(2a)#

The #y#-value can be found out by solving the equation.

So now, we have to expand the equation we have to get it in quadratic form:

#5(x+3)^2-9=y#

#-> 5(x+3)(x+3)-9=y#

#-> 5(x^2+6x+9)-9=y#

#-> 5x^2+30x+45-9=y#

#-> 5x^2+30x+36=y#

Now, #a=5# and #b=30#. (FYI, #c=36#)

#-> (-b)/(2a)=(-(30))/(2(5))#

#->(-b)/(2a) = (-30)/10#

#->(-b)/(2a) = -3#

Thus, the #x#-value #=-3#. Now, we substitute #-3# for #x# to get the #y# value of the vertex:

#5x^2+30x+36=y#

becomes:

#5(-3)^2+30(-3)+36=y#

#-> 45+(-90)+36=y#

#-> y=81-90#

#-> y=-9#

Thus, since #x=-3# and #y=-9#, the vertex is:

#(-3 , -9)#

2) This is the easier way of doing it - by using the Vertex Formula:

In the equation #a(x-h)^2+k=y#, the vertex is #(h,k)#

We are already given an equation in the Vertex format, so it is easy to find out the Vertex coordinates:

#5(x+3)^2-9=y#

can be rewritten as:

#5(x-(-3))^2-9=y#

Now we have it in the Vertex-form, where #h=-3#, and #k=-9#

So, the Vertex coordinates are:

#(h,k)#

#=(-3,-9)#

Tip: you can change an equation in a quadratic form to a vertex form by completing the square. If you are not aware of this concept, search it up on the Internet or post a question on Socratic.