What is the vertex of #y = 2(x - 3)^2 + 1#?

1 Answer
Mar 8, 2017

Answer:

#(3,1)#

Explanation:

The answer happens to be in the equation without any work needed!

This equation is already in vertex form. The equation for vertex form is #y=a(x-h)^2+k#

In this equation, #(h,k)# is the vertex. When you look at your equation: #y=2(x-3)^2+1#, the #h,k# is #(3,1)#.

*Note: in vertex form, there is a negative sign in front of the #h#, so you must take whatever h value there is an make is opposite. Thus, your vertex is #(3,1)# and not #(-3,1)#.

Check your answer with a graph!
graph{2(x-3)^2+1 [-3.354, 6.646, -1.18, 3.82]}