How do you find the vertex of #f(x)=3(x+4)^2+2#?

1 Answer
Dec 18, 2017

Compare with the general vertex form to get that the vertex is at #(-4, 2)#.

Explanation:

That function, #f(x) = 3(x + 4)^2 + 2# is already in the vertex form, #f(x) = a(x - h) + k#.

We could then see that #k#, the #y#-coordinate of the vertex, is #2#. However, for the #x#-coordinate, in the vertex form it should be subtracted, but in our function it is added.

No worries! Since two subtractions "combine" into an addition, we could reverse this process:

#f(x) = 3(x + 4)^2 + 2 rarr f(x) = 3(x - (- 4))^2 + 2#

Now we can compare with the vertex form to see that #h#, the #x#-coordinate of the vertex, is at #-4#.

To conclude, the vertex is at #(-4, 2)#. Here's what the graph looks like:

graph{y=3(x+4)^2+2 [-10, 10, -5, 5]}