What are the coordinates of the vertex of the parabola whose equation is #y = 3(x - 2)^2 + 5#?

Question

2413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-12T15:16:44

The answer is: #V(2,5)#.

There are two ways.

First:

we can remember the equation of the parabola, given the vertex #V(x_v,y_v)# and the amplitude #a#:

#y-y_v=a(x-x_v)^2#.

So:

#y-5=3(x-2)^2# has vertex: #V(2,5)#.

Second:

we can make the counts:

#y=3(x^2-4x+4)+5rArry=3x^2-12x+17#

and, remembering that #V(-b/(2a),-Delta/(4a))#,

#V(-(-12)/(2*3),-(12^2-4*3*17)/(4*3))rArrV(2,5)#.

Answer 2 · 2015-04-12T16:00:01

Vertex is #(2, 5)#

Method

Use the form: #(x - h)^2 = 4a(y - k)#

This parabola has vertex at #(h, k)#
And its major axis is along the #y-"axis"#

In our case we have, #y = 3(x - 2)^2 + 5#

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

#=> (x - 2)^2 = 1/3(y - 5)#

So, the vertex is #(2, 5)#

Worthy of note
When the equation is of the form: #(y - k)^2 = 4a(x - h)#

The vertex is at #(h, k)# and the parabola lies along the #x-"axis"#