What are the vertices of the graph given by the equation #(x+6)^2/4 = 1#?

1 Answer
Jun 6, 2018

Answer:

I think there's something wrong with the question, please see below.

Explanation:

Expanding your expression gives

#\frac{(x+6)^2}{4}=1#

#\therefore (x+6)^2 = 4#

#\therefore x^2+12x+36 = 4#

#\therefore x^2+12x+32 = 0#

This is not really the equation of something you can graph, since a graph represents a relation between the #x# values and the #y# values (or however, in general, the relation between an indipendent variable and a dependent one).

In this case, we only have one variable, and the equation equals zero. The best we can do in this case is to solve the equation, i.e. to find the values of #x# that satisfy the equation. In this case, the solutions are #x=-8# and #x=-4#.